pppp

"Hello, World!" Program
In this example, you will learn to print "Hello, World!" on the screen in C programming.
Display "Hello, World!"
#include <stdio.h>
int main() {
// printf() displays the string inside quotation
printf("Hello, World!");
return 0;
}

Output 
Hello, World!
How "Hello, World!" program works?
The #include is a preprocessor command that tells the compiler to include the contents of stdio.h (standard input and output) file in the program.
The stdio.h file contains functions such as scanf() and printf() to take input and display output respectively.
If you use the printf() function without writing #include <stdio.h>, the program will not compile.
The execution of a C program starts from the main() function.
printf() is a library function to send formatted output to the screen. In this program, printf() displays Hello, World! text on the screen.
The return 0; statement is the "Exit status" of the program. In simple terms, the program ends with this statement.
Check Whether a Number is Prime or Not
In this example, you will learn to check whether an integer entered by the user is a prime number or not.
A prime number is a positive integer that is divisible only by 1 and itself. For example: 2, 3, 5, 7, 11, 13, 17
Check Prime Number
#include <stdio.h>
int main() {
int n, i, flag = 0;
printf("Enter a positive integer: ");
scanf("%d", &n);
for (i = 2; i <= n / 2; ++i) {
// condition for non-prime
if (n % i == 0) {
flag = 1;
break;
}
}
if (n == 1) {
printf("1 is neither prime nor composite.");
} 
else {
if (flag == 0)
printf("%d is a prime number.", n);
else
printf("%d is not a prime number.", n);
}
return 0;
}
Output
Enter a positive integer: 29
29 is a prime number.
In the program, a for loop is iterated from i = 2 to i < n/2.
In each iteration, whether n is perfectly divisible by i is checked using:
if (n % i == 0) {
flag = 1;
    break;
}
If n is perfectly divisible by i, n is not a prime number. In this case, flag is set to 1, and the loop is terminated using the break statement.
Notice that we have initialized flag as 0 during the start of our program.
So, if n is a prime number after the loop, flag will still be 0. However, if n is a non-prime number, flag will be 1.
Visit this page to learn how you can print all the prime numbers between two intervals.
Print an Integer (Entered by the User)
In this example, the integer entered by the user is stored in a variable and printed on the screen.
Print an Integer
#include <stdio.h>
int main() {   
int number;
printf("Enter an integer: ");  
// reads and stores input
scanf("%d", &number);
// displays output
printf("You entered: %d", number);
return 0;
}

Output 
Enter an integer: 25
You entered: 25
In this program, an integer variable number is declared.
int number;
Then, the user is asked to enter an integer number. This number is stored in the number variable.
printf("Enter an integer: ");
scanf("%d", &number);
Finally, the value stored in number is displayed on the screen using printf().
printf("You entered: %d", number);
Add Two Integers
In this example, the user is asked to enter two integers. Then, the sum of these two integers is calculated and displayed on the screen.
Add Two Integers
#include <stdio.h>
int main() {    
int number1, number2, sum;
printf("Enter two integers: ");
scanf("%d %d", &number1, &number2);
// calculating sum
sum = number1 + number2;      
printf("%d + %d = %d", number1, number2, sum);
return 0;
}
Output
Enter two integers: 12
11
12 + 11 = 23
In this program, the user is asked to enter two integers. These two integers are stored in variables number1 and number2 respectively.
printf("Enter two integers: ");
scanf("%d %d", &number1, &number2);
Then, these two numbers are added using the + operator, and the result is stored in the sum variable.
sum = number1 + number2;


Finally, the printf() function is used to display the sum of numbers.
printf("%d + %d = %d", number1, number2, sum);
Multiply Two Floating-Point Numbers
In this example, the product of two floating-point numbers entered by the user is calculated and printed on the screen.
Multiply Two Numbers
#include <stdio.h>
int main() {
double a, b, product;
printf("Enter two numbers: ");
scanf("%lf %lf", &a, &b);  
// Calculating product
product = a * b;
// %.2lf displays number up to 2 decimal point
printf("Product = %.2lf", product);
return 0;
}

Output 
Enter two numbers: 2.4
1.12
Product = 2.69
In this program, the user is asked to enter two numbers which are stored in variables a and b respectively.
printf("Enter two numbers: ");
scanf("%lf %lf", &a, &b); 
Then, the product of a and b is evaluated and the result is stored in product.
product = a * b;
Finally, product is displayed on the screen using printf().
printf("Product = %.2lf", product);
Notice that, the result is rounded off to the second decimal place using %.2lf conversion character.
Find ASCII Value of a Character
In this example, you will learn how to find the ASCII value of a character.
In C programming, a character variable holds ASCII value (an integer number between 0 and 127) rather than that character itself. This integer value is the ASCII code of the character.
For example, the ASCII value of 'A' is 65.
What this means is that, if you assign 'A' to a character variable, 65 is stored in the variable rather than 'A' itself.
Now, let's see how we can print the ASCII value of characters in C programming.
Print ASCII Value
#include <stdio.h>
int main() {  
char c;
printf("Enter a character: ");
scanf("%c", &c);  
// %d displays the integer value of a character
// %c displays the actual character
printf("ASCII value of %c = %d", c, c);
return 0;
}
Output
Enter a character: G
ASCII value of G = 71
In this program, the user is asked to enter a character. The character is stored in variable c.
When %d format string is used, 71 (the ASCII value of G) is displayed.
When %c format string is used, 'G' itself is displayed.
Compute Quotient and Remainder 
In this example, you will learn to find the quotient and remainder when an integer is divided by another integer.
Compute Quotient and Remainder
#include <stdio.h>
int main() {
int dividend, divisor, quotient, remainder;
printf("Enter dividend: ");
scanf("%d", &dividend);
printf("Enter divisor: ");
scanf("%d", &divisor);
// Computes quotient
quotient = dividend / divisor;
// Computes remainder
remainder = dividend % divisor;
printf("Quotient = %d\n", quotient);
printf("Remainder = %d", remainder);
return 0;
}

Output 
Enter dividend: 25
Enter divisor: 4
Quotient = 6
Remainder = 1 
In this program, the user is asked to enter two integers (dividend and divisor). They are stored in variables dividend and divisor respectively.
printf("Enter dividend: ");
scanf("%d", &dividend);
printf("Enter divisor: ");
scanf("%d", &divisor);
Then the quotient is evaluated using / (the division operator), and stored in quotient.
quotient = dividend / divisor;
Similarly, the remainder is evaluated using % (the modulo operator) and stored in remainder.
remainder = dividend % divisor;
Finally, the quotient and remainder are displayed using printf().
printf("Quotient = %d\n", quotient);
printf("Remainder = %d", remainder);
Learn more about how operators work in C programming.
Find the Size of int, float, double and char
In this example, you will learn to evaluate the size of each variable using sizeof operator.
The sizeof(variable) operator computes the size of a variable. And, to print the result returned by sizeof, we use either %lu or %zu format specifier.
Find the Size of Variables
#include<stdio.h>
int main() {
int intType;
float floatType;
double doubleType;
char charType;
// sizeof evaluates the size of a variable
printf("Size of int: %zu bytes\n", sizeof(intType));
printf("Size of float: %zu bytes\n", sizeof(floatType));
printf("Size of double: %zu bytes\n", sizeof(doubleType));
printf("Size of char: %zu byte\n", sizeof(charType));
return 0;
}

Output 
Size of int: 4 bytes
Size of float: 4 bytes
Size of double: 8 bytes
Size of char: 1 byte
In this program, 4 variables intType, floatType, doubleType and charType are declared.
Then, the size of each variable is computed using the sizeof operator.
Demonstrate the Working of Keyword long
In this example, you will learn to demonstrate the working of the long keyword.
Program Using the long keyword
#include <stdio.h>
int main() {
int a;
long b;   // equivalent to long int b;
long long c;  // equivalent to long long int c;
double e;
long double f;
printf("Size of int = %zu bytes \n", sizeof(a));
printf("Size of long int = %zu bytes\n", sizeof(b));
printf("Size of long long int = %zu bytes\n", sizeof(c));
printf("Size of double = %zu bytes\n", sizeof(e));
printf("Size of long double = %zu bytes\n", sizeof(f));
return 0;
}
Output
Size of int = 4 bytes 
Size of long int = 8 bytes
Size of long long int = 8 bytes
Size of double = 8 bytes
Size of long double = 16 bytes
In this program, the sizeof operator is used to find the size of int, long, long long, double and long double variables.
As you can see, the size of long int and long double variables are larger than int and double variables, respectively.
By the way, the sizeof operator returns size_t (unsigned integral type).
The size_t data type is used to represent the size of an object. The format specifier used for size_t is %zu.
Note: The long keyword cannot be used with float and char types.
Swap Two Numbers
In this example, you will learn to swap two numbers in C programming using two different techniques. 
Swap Numbers Using Temporary Variable
#include<stdio.h>
int main() {
double first, second, temp;
printf("Enter first number: ");
scanf("%lf", &first);
printf("Enter second number: ");
scanf("%lf", &second);
// Value of first is assigned to temp
temp = first;
// Value of second is assigned to first
first = second;
// Value of temp (initial value of first) is assigned to second
second = temp;
      // %.2lf displays number up to 2 decimal points
printf("\nAfter swapping, firstNumber = %.2lf\n", first);
printf("After swapping, secondNumber = %.2lf", second);
return 0;
}


Output 
Enter first number: 1.20
Enter second number: 2.45
After swapping, firstNumber = 2.45
After swapping, secondNumber = 1.20
In the above program, the temp variable is assigned the value of the first variable.
Then, the value of the first variable is assigned to the second variable.
Finally, the temp (which holds the initial value of first) is assigned to second. This completes the swapping process.
Swap Numbers Without Using Temporary Variables
#include <stdio.h>
int main() {
double a, b;
printf("Enter a: ");
scanf("%lf", &a);
printf("Enter b: ");
scanf("%lf", &b);
// Swapping
    // a = (initial_a - initial_b)
a = a - b;   
 
    // b = (initial_a - initial_b) + initial_b = initial_a
b = a + b;
    // a = initial_a - (initial_a - initial_b) = initial_b
a = b - a;
​​​​    ​// %.2lf displays number up to 2 decimal points
printf("After swapping, a = %.2lf\n", a);
printf("After swapping, b = %.2lf", b);
return 0;
}
Output
Enter a: 10.25
Enter b: -12.5
After swapping, a = -12.50
After swapping, b = 10.25
Check Whether a Number is Even or Odd
In this example, you will learn to check whether a number entered by the user is even or odd.
An even number is an integer that is exactly divisible by 2. For example: 0, 8, -24
An odd number is an integer that is not exactly divisible by 2. For example: 1, 7, -11, 15
Check Even or Odd
#include <stdio.h>
int main() {
int num;
printf("Enter an integer: ");
scanf("%d", &num);
// true if num is perfectly divisible by 2
if(num % 2 == 0)
printf("%d is even.", num);
else
printf("%d is odd.", num);
return 0;
}

Output 
Enter an integer: -7
-7 is odd.
In the program, the integer entered by the user is stored in the variable num.
Then, whether num is perfectly divisible by 2 or not is checked using the modulus % operator.
If the number is perfectly divisible by 2, test expression number%2 == 0 evaluates to 1 (true). This means the number is even.
However, if the test expression evaluates to 0 (false), the number is odd.
Check Odd or Even Using the Ternary Operator
#include <stdio.h>
int main() {
int num;
printf("Enter an integer: ");
scanf("%d", &num);
(num % 2 == 0) ? printf("%d is even.", num) : printf("%d is odd.", num);
return 0;
}


Output 
Enter an integer: 33
33 is odd.
In the above program, we have used the ternary operator ?: instead of the if...else statement.
Check Whether a Character is a Vowel or Consonant
In this example, you will learn to check whether an alphabet entered by the user is a vowel or a consonant.
The five letters A, E, I, O and U are called vowels. All other alphabets except these 5 vowels are called consonants.
This program assumes that the user will always enter an alphabet character.
Check Vowel or consonant
#include <stdio.h>
int main() {
char c;
int lowercase_vowel, uppercase_vowel;
printf("Enter an alphabet: ");
scanf("%c", &c);
// evaluates to 1 if variable c is a lowercase vowel
lowercase_vowel = (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
// evaluates to 1 if variable c is a uppercase vowel
uppercase_vowel = (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
// evaluates to 1 (true) if c is a vowel
if (lowercase_vowel || uppercase_vowel)
printf("%c is a vowel.", c);
else
printf("%c is a consonant.", c);
return 0;
}

Output 
Enter an alphabet: G
G is a consonant.
The character entered by the user is stored in variable c.
The lowercase_vowel variable evaluates to 1 (true) if c is a lowercase vowel and 0 (false) for any other characters.
Similarly, the uppercase_vowel variable evaluates to 1 (true) if c is an uppercase vowel and 0 (false) for any other character.
If either lowercase_vowel or uppercase_vowel variable is 1 (true), the entered character is a vowel. However, if both lowercase_vowel and uppercase_vowel variables are 0, the entered character is a consonant.
Note: This program assumes that the user will enter an alphabet. If the user enters a non-alphabetic character, it displays the character is a consonant.
To fix this, we can use the isalpha() function. The islapha() function checks whether a character is an alphabet or not.
#include <ctype.h>
#include <stdio.h>
int main() {
char c;
int lowercase_vowel, uppercase_vowel;
printf("Enter an alphabet: ");
scanf("%c", &c);
// evaluates to 1 if variable c is a lowercase vowel
lowercase_vowel = (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
// evaluates to 1 if variable c is a uppercase vowel
uppercase_vowel = (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
// Show error message if c is not an alphabet
if (!isalpha(c))
printf("Error! Non-alphabetic character.");
else if (lowercase_vowel || uppercase_vowel)
printf("%c is a vowel.", c);
else
printf("%c is a consonant.", c);
return 0;
}
Now, if the user enters a non-alphabetic character, you will see:
Enter an alphabet: 3
Error! Non-alphabetic character.
Find the Largest Number Among Three Numbers
In this example, you will learn to find the largest number among the three numbers entered by the user.
Example 1: Using if Statement
#include <stdio.h>
int main() {
double n1, n2, n3;
printf("Enter three different numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
// if n1 is greater than both n2 and n3, n1 is the largest
if (n1 >= n2 && n1 >= n3)
printf("%.2f is the largest number.", n1);
// if n2 is greater than both n1 and n3, n2 is the largest
if (n2 >= n1 && n2 >= n3)
printf("%.2f is the largest number.", n2);
// if n3 is greater than both n1 and n2, n3 is the largest
if (n3 >= n1 && n3 >= n2)
printf("%.2f is the largest number.", n3);
return 0;
}
Example 2: Using if...else Ladder
#include <stdio.h>
int main() {
double n1, n2, n3;
printf("Enter three numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
// if n1 is greater than both n2 and n3, n1 is the largest
if (n1 >= n2 && n1 >= n3)
printf("%.2lf is the largest number.", n1);
// if n2 is greater than both n1 and n3, n2 is the largest
else if (n2 >= n1 && n2 >= n3)
printf("%.2lf is the largest number.", n2);
// if both above conditions are false, n3 is the largest
else
printf("%.2lf is the largest number.", n3);
return 0;
}
Example 3: Using Nested if...else
#include <stdio.h>
int main() {
double n1, n2, n3;
printf("Enter three numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
if (n1 >= n2) {
if (n1 >= n3)
printf("%.2lf is the largest number.", n1);
else
printf("%.2lf is the largest number.", n3);
} else {
if (n2 >= n3)
printf("%.2lf is the largest number.", n2);
else
printf("%.2lf is the largest number.", n3);
}
return 0;
}
The output of all these programs above will be the same.
Enter three numbers: -4.5
3.9
5.6
5.60 is the largest number.
Find the Roots of a Quadratic Equation
In this example, you will learn to find the roots of a quadratic equation in C programming.
The standard form of a quadratic equation is:
ax2 + bx + c = 0, where
a, b and c are real numbers and
a != 0
The term b2-4ac is known as the discriminant of a quadratic equation. It tells the nature of the roots.
If the discriminant is greater than 0, the roots are real and different.
If the discriminant is equal to 0, the roots are real and equal.
If the discriminant is less than 0, the roots are complex and different.

	

Find Roots of a Quadratic Equation
#include <math.h>
#include <stdio.h>
int main() {
double a, b, c, discriminant, root1, root2, realPart, imagPart;
printf("Enter coefficients a, b and c: ");
scanf("%lf %lf %lf", &a, &b, &c);
discriminant = b * b - 4 * a * c;
// condition for real and different roots
if (discriminant > 0) {
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("root1 = %.2lf and root2 = %.2lf", root1, root2);
}
// condition for real and equal roots
else if (discriminant == 0) {
root1 = root2 = -b / (2 * a);
printf("root1 = root2 = %.2lf;", root1);
}
// if roots are not real
else {
realPart = -b / (2 * a);
imagPart = sqrt(-discriminant) / (2 * a);
printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi", realPart, imagPart, realPart, imagPart);
}
return 0;
} 
Output
Enter coefficients a, b and c: 2.3
4
5.6
root1 = -0.87+1.30i and root2 = -0.87-1.30i
In this program, the sqrt() library function is used to find the square root of a number. To learn more, visit: sqrt() function.
Check Leap Year
In this example, you will learn to check whether the year entered by the user is a leap year or not.
A leap year is exactly divisible by 4 except for century years (years ending with 00). The century year is a leap year only if it is perfectly divisible by 400.
For example,
1999 is not a leap year
2000 is a leap year
2004 is a leap year
Check Leap Year
#include <stdio.h>
int main() {
int year;
printf("Enter a year: ");
scanf("%d", &year);
// leap year if perfectly divisible by 400
if (year % 400 == 0) {
printf("%d is a leap year.", year);
}
// not a leap year if divisible by 100
// but not divisible by 400
else if (year % 100 == 0) {
printf("%d is not a leap year.", year);
}
// leap year if not divisible by 100
// but divisible by 4
else if (year % 4 == 0) {
printf("%d is a leap year.", year);
}
// all other years are not leap years
else {
printf("%d is not a leap year.", year);
}
return 0;
}
Output 1 
Enter a year: 1900
1900 is not a leap year.
Output 2 
Enter a year: 2012
2012 is a leap year.
Check Whether a Number is Positive or Negative
In this example, you will learn to check whether a number (entered by the user) is negative or positive.
This program takes a number from the user and checks whether that number is either positive or negative or zero.
Check Positive or Negative Using if...else
#include <stdio.h>
int main() {
double num;
printf("Enter a number: ");
scanf("%lf", &num);
if (num <= 0.0) {
if (num == 0.0)
printf("You entered 0.");
else
printf("You entered a negative number.");
} else
printf("You entered a positive number.");
return 0;
}
You can also solve this problem using nested if else statement.
Check Positive or Negative Using Nested if...else
#include <stdio.h>
int main() {
double num;
printf("Enter a number: ");
scanf("%lf", &num);
if (num < 0.0)
printf("You entered a negative number.");
else if (num > 0.0)
printf("You entered a positive number.");
else
printf("You entered 0.");
return 0;
}

Output 1
Enter a number: 12.3
You entered a positive number.Output 2
Enter a number: 0
You entered 0.  
Check Whether a Character is an Alphabet or not
In this example, you will learn to check whether a character entered by the user is an alphabet or not.
In C programming, a character variable holds an ASCII value (an integer number between 0 and 127) rather than that character itself.
The ASCII value of the lowercase alphabet is from 97 to 122. And, the ASCII value of the uppercase alphabet is from 65 to 90.
If the ASCII value of the character entered by the user lies in the range of 97 to 122 or from 65 to 90, that number is an alphabet.
Check Alphabet
#include <stdio.h>
int main() {
char c;
printf("Enter a character: ");
scanf("%c", &c);
if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z'))
printf("%c is an alphabet.", c);
else
printf("%c is not an alphabet.", c);
return 0;
}
Output 
Enter a character: *
* is not an alphabet
In the program, 'a' is used instead of 97 and 'z' is used instead of 122. Similarly, 'A' is used instead of 65 and 'Z' is used instead of 90.
	Note: It is recommended we use the isalpha() function to check whether a character is an alphabet or not.
if (isalpha(c))
printf("%c is an alphabet.", c);
else
printf("%c is not an alphabet.", c);
Calculate the Sum of Natural Numbers
In this example, you will learn to calculate the sum of natural numbers entered by the user.
The positive numbers 1, 2, 3... are known as natural numbers. The sum of natural numbers up to 10 is:
sum = 1 + 2 + 3 + ... + 10
Sum of Natural Numbers Using for Loop
#include <stdio.h>
int main() {
int n, i, sum = 0;
printf("Enter a positive integer: ");
scanf("%d", &n);
for (i = 1; i <= n; ++i) {
sum += i;
}
printf("Sum = %d", sum);
return 0;
}

The above program takes input from the user and stores it in the variable n. Then, for loop is used to calculate the sum up to n.
Sum of Natural Numbers Using while Loop
#include <stdio.h>
int main() {
int n, i, sum = 0;
printf("Enter a positive integer: ");
scanf("%d", &n);
i = 1;
while (i <= n) {
sum += i;
++i;
}
printf("Sum = %d", sum);
return 0;
}
Output 
Enter a positive integer: 100
Sum = 5050
In both programs, the loop is iterated n number of times. And, in each iteration, the value of i is added to sum and i is incremented by 1.
Though both programs are technically correct, it is better to use for loop in this case. It's because the number of iterations is known.
The above programs don't work properly if the user enters a negative integer. Here is a little modification to the above program where we keep taking input from the user until a positive integer is entered.
Read Input Until a Positive Integer is Entered
#include <stdio.h>
int main() {
int n, i, sum = 0;
do {
printf("Enter a positive integer: ");
scanf("%d", &n);
} while (n <= 0);
for (i = 1; i <= n; ++i) {
sum += i;
}
printf("Sum = %d", sum);
return 0;
}

Visit this page to learn how to find the sum of natural numbers using recursion.
Find Factorial of a Number
In this example, you will learn to calculate the factorial of a number entered by the user.
The factorial of a positive number n is given by:
factorial of n (n!) = 1 * 2 * 3 * 4....n
The factorial of a negative number doesn't exist. And, the factorial of 0 is
1. 
Factorial of a Number
#include <stdio.h>
int main() {
int n, i;
unsigned long long fact = 1;
printf("Enter an integer: ");
scanf("%d", &n);
// shows error if the user enters a negative integer
if (n < 0)
printf("Error! Factorial of a negative number doesn't exist.");
else {
for (i = 1; i <= n; ++i) {
fact *= i;
}
printf("Factorial of %d = %llu", n, fact);
}
return 0;
}

Output 
Enter an integer: 10
Factorial of 10 = 3628800
This program takes a positive integer from the user and computes the factorial
using for loop.
Since the factorial of a number may be very large, the type of factorial
variable is declared as unsigned long long.
If the user enters a negative number, the program displays a custom error
message.
You can also find the
factorial of a number using recursion.
Generate Multiplication Table
In this example, you will learn to generate the multiplication table of a number entered by the user.
The program below takes an integer input from the user and generates the
multiplication tables up to 10.
Multiplication Table Up to 10
#include <stdio.h>
int main() {
int n, i;
printf("Enter an integer: ");
scanf("%d", &n);
for (i = 1; i <= 10; ++i) {
printf("%d * %d = %d \n", n, i, n * i);
}
return 0;
}

Output 
Enter an integer: 9
9 * 1 = 9
9 * 2 = 18
9 * 3 = 27
9 * 4 = 36
9 * 5 = 45
9 * 6 = 54
9 * 7 = 63
9 * 8 = 72
9 * 9 = 81
9 * 10 = 90
Here's a little modification of the above program to generate the
multiplication table up to a range (where range is also a positive
integer entered by the user).
Multiplication Table Up to a range
#include <stdio.h>
int main() {
int n, i, range;
printf("Enter an integer: ");
scanf("%d", &n);
printf("Enter the range: ");
scanf("%d", &range);
for (i = 1; i <= range; ++i) {
printf("%d * %d = %d \n", n, i, n * i);
}
return 0;
}


Output 
Enter an integer: 12
Enter the range: 8
12 * 1 = 12 
12 * 2 = 24 
12 * 3 = 36 
12 * 4 = 48 
12 * 5 = 60 
12 * 6 = 72 
12 * 7 = 84 
12 * 8 = 96 
Display Fibonacci Sequence
In this example, you will learn to display the Fibonacci sequence of first n numbers (entered by the user).
The Fibonacci sequence is a sequence where the next term is the sum of the previous two terms. The first two terms of the Fibonacci sequence are 0 followed by 1.
The Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21
Visit this page to learn about the Fibonacci sequence.
Fibonacci Series up to n terms
#include <stdio.h>
int main() {
int i, n, t1 = 0, t2 = 1;
    int nextTerm = t1 + t2;
printf("Enter the number of terms: ");
scanf("%d", &n);
printf("Fibonacci Series: %d, %d, ", t1, t2);
for (i = 1; i <= n; ++i) {
printf("%d, ", nextTerm);
t1 = t2;
t2 = nextTerm;
nextTerm = t1 + t2;
}
return 0;
}

Output 
Enter the number of terms: 10
Fibonacci Series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 
Fibonacci Sequence Up to a Certain Number
#include <stdio.h>
int main() {
int t1 = 0, t2 = 1, nextTerm = 0, n;
printf("Enter a positive number: ");
scanf("%d", &n);
// displays the first two terms which is always 0 and 1
printf("Fibonacci Series: %d, %d, ", t1, t2);
nextTerm = t1 + t2;
while (nextTerm <= n) {
printf("%d, ", nextTerm);
t1 = t2;
t2 = nextTerm;
nextTerm = t1 + t2;
}
return 0;
}
Output 
Enter a positive integer: 100
Fibonacci Series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 
 C Program to Find GCD of two Numbers
Examples on different ways to calculate GCD of two integers (for both positive and negative integers) using loops and decision making statements.
The HCF or GCD of two integers is the largest integer that can exactly divide both numbers (without a remainder).
There are many ways to find the greatest common divisor in C programming.
Example #1: GCD Using for loop and if Statement
#include <stdio.h>
int main()
{
int n1, n2, i, gcd;
printf("Enter two integers: ");
scanf("%d %d", &n1, &n2);
for(i=1; i <= n1 && i <= n2; ++i)
{
// Checks if i is factor of both integers
if(n1%i==0 && n2%i==0)
gcd = i;
}
printf("G.C.D of %d and %d is %d", n1, n2, gcd);
return 0;
}
In this program, two integers entered by the user are stored in variable n1 and n2.Then, for loop is iterated until i is less than n1 and n2.
In each iteration, if both n1 and n2 are exactly divisible by i, the value of i is assigned to gcd.
When the for loop is completed, the greatest common divisor of two numbers is stored in variable gcd.
Example #2: GCD Using while loop and if...else Statement
#include <stdio.h>
int main()
{
int n1, n2;
printf("Enter two positive integers: ");
scanf("%d %d",&n1,&n2);
while(n1!=n2)
{
if(n1 > n2)
n1 -= n2;
else
n2 -= n1;
}
printf("GCD = %d",n1);
return 0;
}
Output
Enter two positive integers: 81
153
GCD = 9
This is a better way to find the GCD. In this method, smaller integer is subtracted from the larger integer, and the result is assigned to the variable holding larger integer. This process is continued until n1 and n2 are equal.
The above two programs works as intended only if the user enters positive integers. Here's a little modification of the second example to find the GCD for both positive and negative integers.
Example #3: GCD for both positive and negative numbers
#include <stdio.h>
int main()
{
int n1, n2;
printf("Enter two integers: ");
scanf("%d %d",&n1,&n2);
// if user enters negative number, sign of the number is changed to positive
n1 = ( n1 > 0) ? n1 : -n1;
n2 = ( n2 > 0) ? n2 : -n2;
while(n1!=n2)
{
if(n1 > n2)
n1 -= n2;
else
n2 -= n1;
}
printf("GCD = %d",n1);
return 0;
}
Output
Enter two integers: 81
-153
GCD = 9
You can also use recursion to find the GCD.
Find LCM of two Numbers
In this example, you will learn to calculate the LCM (Lowest common multiple) of two numbers entered by the user.
The LCM of two integers n1 and n2 is the smallest positive integer that is perfectly divisible by both n1 and n2 (without a remainder). For example, the LCM of 72 and 120 is 360.
LCM using while and if
#include <stdio.h>
int main() {
int n1, n2, max;
printf("Enter two positive integers: ");
scanf("%d %d", &n1, &n2);
// maximum number between n1 and n2 is stored in min
max = (n1 > n2) ? n1 : n2;
while (1) {
if (max % n1 == 0 && max % n2 == 0) {
printf("The LCM of %d and %d is %d.", n1, n2, max);
break;
}
++max;
}
return 0;
}

Output 
Enter two positive integers: 72
120
The LCM of 72 and 120 is 360.
In this program, the integers entered by the user are stored in variable n1 and n2 respectively.
The largest number among n1 and n2 is stored in max. The LCM of two numbers cannot be less than max.
The test expression of while loop is always true.
In each iteration, whether max is perfectly divisible by n1 and n2 is checked.
if (min % n1 == 0 && max% n2 == 0) { ... }
If this test condition is not true, max is incremented by 1 and the iteration continues until the test expression of the if statement is true.
The LCM of two numbers can also be found using the formula:
LCM = (num1*num2)/GCD
Learn how to find the GCD of two numbers in C programming.
LCM Calculation Using GCD
#include <stdio.h>
int main() {
int n1, n2, i, gcd, lcm;
printf("Enter two positive integers: ");
scanf("%d %d", &n1, &n2);
for (i = 1; i <= n1 && i <= n2; ++i) {
// check if i is a factor of both integers
if (n1 % i == 0 && n2 % i == 0)
gcd = i;
}
lcm = (n1 * n2) / gcd;
printf("The LCM of two numbers %d and %d is %d.", n1, n2, lcm);
return 0;
}

Output 
Enter two positive integers: 72
120
The LCM of two numbers 72 and 120 is 360.
Display Characters from A to Z Using Loop
In this example, you will learn to print all the letters of the English alphabet.
Print English Alphabets
#include <stdio.h>
int main() {
char c;
for (c = 'A'; c <= 'Z'; ++c)
printf("%c ", c);
return 0;
}

Output 
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
In this program, the for loop is used to display the English
alphabet in uppercase.
Here's a little modification of the above program. The program displays the
English alphabet in either uppercase or lowercase depending upon the input
given by the user.
Print Lowercase/Uppercase alphabets
#include <stdio.h>
int main() {
char c;
printf("Enter u to display uppercase alphabets.\n");
printf("Enter l to display lowercase alphabets. \n");
scanf("%c", &c);
if (c == 'U' || c == 'u') {
for (c = 'A'; c <= 'Z'; ++c)
printf("%c ", c);
} else if (c == 'L' || c == 'l') {
for (c = 'a'; c <= 'z'; ++c)
printf("%c ", c);
} else {
printf("Error! You entered an invalid character.");
}
return 0;
}


Output 
Enter u to display uppercase alphabets. 
Enter l to display lowercase alphabets. l
a b c d e f g h i j k l m n o p q r s t u v w x y z
Count Number of Digits in an Integer
In this example, you will learn to count the number of digits in an integer entered by the user.
This program takes an integer from the user and calculates the number of digits. For example: If the user enters 2319, the output of the program will be 4.
Count the Number of Digits
#include <stdio.h>
int main() {
long long n;
int count = 0;
printf("Enter an integer: ");
scanf("%lld", &n);
// iterate until n becomes 0
// remove last digit from n in each iteration
// increase count by 1 in each iteration
while (n != 0) {
n /= 10;     // n = n/10
++count;
}
printf("Number of digits: %d", count);
}
Output
Enter an integer: 3452
Number of digits: 4
The integer entered by the user is stored in variable n. Then the while loop is iterated until the test expression n! = 0 is evaluated to 0 (false).
After the first iteration, the value of n will be 345 and the count is incremented to 1.
After the second iteration, the value of n will be 34 and the count is incremented to 2.
After the third iteration, the value of n will be 3 and the count is incremented to 3.
After the fourth iteration, the value of n will be 0 and the count is incremented to 4.
Then the test expression of the loop is evaluated to false and the loop terminates.
Reverse a Number 
In this example, you will learn to reverse the number entered by the user.
Reverse an Integer
#include <stdio.h>
int main() {
int n, rev = 0, remainder;
printf("Enter an integer: ");
scanf("%d", &n);
while (n != 0) {
remainder = n % 10;
rev = rev * 10 + remainder;
n /= 10;
}
printf("Reversed number = %d", rev);
return 0;
}

Output
Enter an integer: 2345
Reversed number = 5432
This program takes an integer input from the user. Then the
while loop is used until n != 0 is false (0).
In each iteration of the loop, the remainder when n is divided by
10 is calculated and the value of n is reduced by 10 times.
Inside the loop, the reversed number is computed using:
rev = rev*10 + remainder;
Calculate the Power of a Number
In this example, you will learn to calculate the power of a number.
The program below takes two integers from the user (a base number and an exponent) and calculates the power.
For example: In the case of 23 
2 is the base number
3 is the exponent
And, the power is equal to 2*2*2
Power of a Number Using the while Loop
#include <stdio.h>
int main() {
int base, exp;
long double result = 1.0;
printf("Enter a base number: ");
scanf("%d", &base);
printf("Enter an exponent: ");
scanf("%d", &exp);
while (exp != 0) {
result *= base;
--exp;
}
printf("Answer = %.0Lf", result);
return 0;
}
Output
Enter a base number: 3
Enter an exponent: 4
Answer = 81
We can also use the pow() function to calculate the power of a number.
Power Using pow() Function
#include <math.h>
#include <stdio.h>
int main() {
double base, exp, result;
printf("Enter a base number: ");
scanf("%lf", &base);
printf("Enter an exponent: ");
scanf("%lf", &exp);
// calculates the power
result = pow(base, exp);
printf("%.1lf^%.1lf = %.2lf", base, exp, result);
return 0;
}
Output
Enter a base number: 2.3
Enter an exponent: 4.5
2.3^4.5 = 42.44
The programs above can only calculate the power of the base number if the exponent is positive. For negative exponents, use the following mathematical logic:
base(-exponent) = 1 / (baseexponent)
For example,
2-3 = 1 / (23)
Check Whether a Number is Palindrome or Not
In this example, you will learn to check whether the number entered by the user is a palindrome or not.
An integer is a palindrome if the reverse of that number is equal to the
original number.
Check Palindrome
#include <stdio.h>
int main() {
int n, reversedN = 0, remainder, originalN;
printf("Enter an integer: ");
scanf("%d", &n);
originalN = n;
// reversed integer is stored in reversedN
while (n != 0) {
remainder = n % 10;
reversedN = reversedN * 10 + remainder;
n /= 10;
}
// palindrome if orignalN and reversedN are equal
if (originalN == reversedN)
printf("%d is a palindrome.", originalN);
else
printf("%d is not a palindrome.", originalN);
return 0;
}


Output 
Enter an integer: 1001
1001 is a palindrome.
Here, the user is asked to enter an integer. The number is stored in variable
n.
We then assigned this number to another variable orignalN. Then,
the reverse of n is found and stored in reversedN.
If originalN is equal to reversedN, the number entered
by the user is a palindrome,
Display Prime Numbers Between Two Intervals
 In this example, you will learn to print all prime numbers between two numbers entered by the user.
Display Prime Numbers Between Two Intervals
#include <stdio.h>
int main() {
int low, high, i, flag;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Prime numbers between %d and %d are: ", low, high);
// iteration until low is not equal to high
while (low < high) {
flag = 0;
// ignore numbers less than 2
if (low <= 1) {
++low;
continue;
}
// if low is a non-prime number, flag will be 1
for (i = 2; i <= low / 2; ++i) {
if (low % i == 0) {
flag = 1;
break;
}
}
if (flag == 0)
printf("%d ", low);
// to check prime for the next number
// increase low by 1
++low;
}
return 0;
}

Output 
Enter two numbers(intervals): 20 
50
Prime numbers between 20 and 50 are: 23 29 31 37 41 43 47
In this program, the while loop is iterated ( high-low-1) times.
In each iteration, whether low is a prime number or not is checked, and the value of low is incremented by 1 until low is equal to high.
Visit this page to learn more about how to check whether a number is prime or not.
If the user enters the larger number first, the above program doesn't work as intended. You can solve this issue by swapping the numbers.
Display Prime Numbers when Larger Number is Entered first
#include <stdio.h>
int main() {
int low, high, i, flag, temp;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
// swap numbers if low is greather than high
if (low > high) {
temp = low;
low = high;
high = temp;
}
printf("Prime numbers between %d and %d are: ", low, high);
while (low < high) {
flag = 0;
// ignore numbers less than 2
if (low <= 1) {
++low;
continue;
}
for (i = 2; i <= low / 2; ++i) {
if (low % i == 0) {
flag = 1;
break;
}
}
if (flag == 0)
printf("%d ", low);
++low;
}
return 0;
}
Visit this page to learn how you can display all the prime numbers between the two intervals by creating a user-defined function
Check Armstrong Number
In this example, you will learn to check whether an integer entered by the user is an Armstrong number or not.
A positive integer is called an Armstrong number (of order n) if
abcd... = an + bn + cn + dn + 
In the case of an Armstrong number of 3 digits, the sum of cubes of each digit is equal to the number itself. For example, 153 is an Armstrong number because
153 = 1*1*1 + 5*5*5 + 3*3*3 
Check Armstrong Number of three digits
#include <stdio.h>
int main() {
int num, originalNum, remainder, result = 0;
printf("Enter a three-digit integer: ");
scanf("%d", &num);
originalNum = num;
while (originalNum != 0) {
// remainder contains the last digit
remainder = originalNum % 10;
result += remainder * remainder * remainder;
// removing last digit from the orignal number
originalNum /= 10;
}
if (result == num)
printf("%d is an Armstrong number.", num);
else
printf("%d is not an Armstrong number.", num);
return 0;
}
Output
Enter a three-digit integer: 371
371 is an Armstrong number.
Check Armstrong Number of n digits
#include <math.h>
#include <stdio.h>
int main() {
int num, originalNum, remainder, n = 0;
float result = 0.0;
printf("Enter an integer: ");
scanf("%d", &num);
originalNum = num;
// store the number of digits of num in n
for (originalNum = num; originalNum != 0; ++n) {
originalNum /= 10;
}
for (originalNum = num; originalNum != 0; originalNum /= 10) {
remainder = originalNum % 10;
// store the sum of the power of individual digits in result
result += pow(remainder, n);
}
// if num is equal to result, the number is an Armstrong number
if ((int)result == num)
printf("%d is an Armstrong number.", num);
else
printf("%d is not an Armstrong number.", num);
return 0;
}

Output 
Enter an integer: 1634
1634 is an Armstrong number.
In this program, the number of digits of an integer is calculated first and stored in n. And, the pow() function is used to compute the power of individual digits in each iteration of the second for loop.
Display Armstrong Number Between Two Intervals
In this example, you will learn to find all Armstrong numbers between two integers entered by the user.
A positive integer is called an Armstrong number (of order n) if
abcd... = an + bn + cn + dn + 
In the case of an Armstrong number of 3 digits, the sum of cubes of each digit is equal to the number itself. For example, 153 is an Armstrong number because
153 = 1*1*1 + 5*5*5 + 3*3*3
Before trying this program, learn how to check whether an integer is an Armstrong number or not.
Armstrong Numbers Between Two Integers
#include <math.h>
#include <stdio.h>
int main() {
int low, high, number, originalNumber, rem, count = 0;
double result = 0.0;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Armstrong numbers between %d and %d are: ", low, high);
// iterate number from (low + 1) to (high - 1)
// In each iteration, check if number is Armstrong
for (number = low + 1; number < high; ++number) {
originalNumber = number;
// number of digits calculation
while (originalNumber != 0) {
originalNumber /= 10;
++count;
}
originalNumber = number;
// result contains sum of nth power of individual digits
while (originalNumber != 0) {
rem = originalNumber % 10;
result += pow(rem, count);
originalNumber /= 10;
}
// check if number is equal to the sum of nth power of individual digits
if ((int)result == number) {
printf("%d ", number);
}
// resetting the values
count = 0;
result = 0;
}
return 0;
}

Output 
Enter two numbers(intervals): 200
2000
Armstrong numbers between 200 and 2000 are: 370 371 407 1634 
In the program, the outer loop is iterated from (low+ 1) to (high - 1). In each iteration, it's checked whether number is an Armstrong number or not.
Inside the outer loop, the number of digits of an integer is calculated first and stored in count. And, the sum of the power of individual digits is stored in the result variable.
If number is equal to result, the number is an Armstrong number.
Note: You need to reset count and result to 0 in each iteration of the outer loop.
Display Factors of a Number
In this example, you will learn to find all the factors of an integer entered by the user.
This program takes a positive integer from the user and displays all the
positive factors of that number.
Factors of a Positive Integer
#include <stdio.h>
int main() {
int num, i;
printf("Enter a positive integer: ");
scanf("%d", &num);
printf("Factors of %d are: ", num);
for (i = 1; i <= num; ++i) {
if (num % i == 0) {
printf("%d ", i);
}
}
return 0;
}
Output
Enter a positive integer: 60
Factors of 60 are: 1 2 3 4 5 6 10 12 15 20 30 60
In the program, a positive integer entered by the user is stored in
num.
The for loop is iterated until i <= num is false.
In each iteration, whether num is exactly divisible by
i is checked. It is the condition for i to be a factor
of num.
if (num % i == 0) {
printf("%d ", i);
}
Then the value of i is incremented by 1.
Make a Simple Calculator Using switch...case
In this example, you will learn to create a simple calculator in C programming using the switch statement.
This program takes an arithmetic operator +, -, *, / and two operands from the user. Then, it performs the calculation on the two operands depending upon the operator entered by the user.
Simple Calculator using switch Statement
#include <stdio.h>
int main() {
char operator;
double first, second;
printf("Enter an operator (+, -, *,): ");
scanf("%c", &operator);
printf("Enter two operands: ");
scanf("%lf %lf", &first, &second);
switch (operator) {
case '+':
printf("%.1lf + %.1lf = %.1lf", first, second, first + second);
break;
case '-':
printf("%.1lf - %.1lf = %.1lf", first, second, first - second);
break;
case '*':
printf("%.1lf * %.1lf = %.1lf", first, second, first * second);
break;
case '/':
printf("%.1lf / %.1lf = %.1lf", first, second, first / second);
break;
// operator doesn't match any case constant
default:
printf("Error! operator is not correct");
}
return 0;
}

Output 
Enter an operator (+, -, *,): *
Enter two operands: 1.5
4.5
1.5 * 4.5 = 6.8
The * operator entered by the user is stored in operator. And, the two operands, 1.5 and 4.5 are stored in first and second respectively.
Since the operator * matches case '*':, the control of the program jumps to
printf("%.1lf * %.1lf = %.1lf", first, second, first * second);
This statement calculates the product and displays it on the screen.
To make our output look cleaner, we have simply limited the output to one decimal place using the code %.1lf.
Finally, the break; statement ends the switch statement.
Display Prime Numbers Between Intervals Using Function
In this example, you will learn to print all prime numbers between two numbers (entered by the user).
To find all the prime numbers between the two integers,
checkPrimeNumber() is created. This function
checks whether a number is prime or not.
Prime Numbers Between Two Integers
#include <stdio.h>
int checkPrimeNumber(int n);
int main() {
int n1, n2, i, flag;
printf("Enter two positive integers: ");
scanf("%d %d", &n1, &n2);
printf("Prime numbers between %d and %d are: ", n1, n2);
for (i = n1 + 1; i < n2; ++i) {
// flag will be equal to 1 if i is prime
flag = checkPrimeNumber(i);
if (flag == 1)
printf("%d ", i);
}
return 0;
}
// user-defined function to check prime number
int checkPrimeNumber(int n) {
int j, flag = 1;
for (j = 2; j <= n / 2; ++j) {
if (n % j == 0) {
flag = 0;
break;
}
}
return flag;
}
Output
Enter two positive integers: 12
30
Prime numbers between 12 and 30 are: 13 17 19 23 29
If the user enters the larger number first, this program will not work as
intended. To solve this issue, you need to swap the numbers first.
Check Prime or Armstrong Number Using User-defined Function
In this example, you will learn to check whether an integer is a prime number or an Armstrong or both by creating two separate functions.
In this program, two user-defined functions checkPrimeNumber() and checkArmstrongNumber() are created.
The checkPrimeNumber() function returns 1 if the number entered by the user is a prime number. Similarly, checkArmstrongNumber() function also returns 1 if the number entered by the user is an Armstrong number.

Check Prime and Armstrong
#include <math.h>
#include <stdio.h>
int checkPrimeNumber(int n);
int checkArmstrongNumber(int n);
int main() {
int n, flag;
printf("Enter a positive integer: ");
scanf("%d", &n);
// check prime number
flag = checkPrimeNumber(n);
if (flag == 1)
printf("%d is a prime number.\n", n);
else
printf("%d is not a prime number.\n", n);
// check Armstrong number
flag = checkArmstrongNumber(n);
if (flag == 1)
printf("%d is an Armstrong number.", n);
else
printf("%d is not an Armstrong number.", n);
return 0;
}
// function to check prime number
int checkPrimeNumber(int n) {
int i, flag = 1, squareRoot;
// computing the square root
squareRoot = sqrt(n);
for (i = 2; i <= squareRoot; ++i) {
// condition for non-prime number
if (n % i == 0) {
flag = 0;
break;
}
}
return flag;
}
// function to check Armstrong number
int checkArmstrongNumber(int num) {
int originalNum, remainder, n = 0, flag;
double result = 0.0;
// store the number of digits of num in n
for (originalNum = num; originalNum != 0; ++n) {
originalNum /= 10;
}
for (originalNum = num; originalNum != 0; originalNum /= 10) {
remainder = originalNum % 10;
// store the sum of the power of individual digits in result
result += pow(remainder, n);
}
// condition for Armstrong number
if (round(result) == num)
flag = 1;
else
flag = 0;
return flag;
}
Output
Enter a positive integer: 407
407 is not a prime number.
407 is an Armstrong number.
Check Whether a Number can be Expressed as Sum of Two Prime Numbers
In this example, you will learn to check if an integer entered by the user can be expressed as the sum of two prime numbers of all possible combinations.
To accomplish this task, we will create a function named
checkPrime().
The checkPrime() returns 1 if the number passed to the function
is a
prime number.
Integer as a Sum of Two Prime Numbers
#include <stdio.h>
int checkPrime(int n);
int main() {
int n, i, flag = 0;
printf("Enter a positive integer: ");
scanf("%d", &n);
for (i = 2; i <= n / 2; ++i) {
// condition for i to be a prime number
if (checkPrime(i) == 1) {
// condition for n-i to be a prime number
if (checkPrime(n - i) == 1) {
printf("%d = %d + %d\n", n, i, n - i);
flag = 1;
}
}
}
if (flag == 0)
printf("%d cannot be expressed as the sum of two prime numbers.", n);
return 0;
}
// function to check prime number
int checkPrime(int n) {
int i, isPrime = 1;
for (i = 2; i <= n / 2; ++i) {
if (n % i == 0) {
isPrime = 0;
break;
}
}
return isPrime;
}
Output
Enter a positive integer: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 17 + 17  
Find the Sum of Natural Numbers using Recursion
In this example, you will learn to find the sum of natural numbers using a recursive function.
The positive numbers 1, 2, 3... are known as natural numbers. The program
below takes a positive integer from the user and calculates the sum up to the
given number.
Visit this page to
find the sum of natural numbers using a loop.
Sum of Natural Numbers Using Recursion
#include <stdio.h>
int addNumbers(int n);
int main() {
int num;
printf("Enter a positive integer: ");
scanf("%d", &num);
printf("Sum = %d", addNumbers(num));
return 0;
}
int addNumbers(int n) {
if (n != 0)
return n + addNumbers(n - 1);
else
return n;
}
Output
Enter a positive integer: 20
Sum = 210
Suppose the user entered 20.
Initially, addNumbers() is called from main() with
20 passed as an argument.
The number 20 is added to the result of addNumbers(19).
In the next function call from addNumbers() to
addNumbers(), 19 is passed which is added to the result of
addNumbers(18). This process continues until n is
equal to 0.
When n is equal to 0, there is no recursive call. This returns the
sum of integers ultimately to the main() function.
  
Find Factorial of a Number Using Recursion
In this example, you will learn to find the factorial of a non-negative integer entered by the user using recursion.
The factorial of a positive number n is given by:
factorial of n (n!) = 1 * 2 * 3 * 4 *...  * n
The factorial of a negative number doesn't exist. And the factorial of
0 is 1.
You will learn to find the factorial of a number using recursion in this
example. Visit this page to learn how you can find the
factorial of a number using a loop.
Factorial of a Number Using Recursion
#include<stdio.h>
long int multiplyNumbers(int n);
int main() {
int n;
printf("Enter a positive integer: ");
scanf("%d",&n);
printf("Factorial of %d = %ld", n, multiplyNumbers(n));
return 0;
}
long int multiplyNumbers(int n) {
if (n>=1)
return n*multiplyNumbers(n-1);
else
return 1;
}

Output 
Enter a positive integer: 6
Factorial of 6 = 720
Suppose the user entered 6.
Initially, multiplyNumbers() is called from
main() with 6 passed as an argument.
Then, 5 is passed to multiplyNumbers() from the same function
(recursive call). In each recursive call, the value of argument
n is decreased by 1.
When the value of n is less than 1, there is no recursive call and
the factorial is returned ultimately to the main() function.
  
Find G.C.D Using Recursion
In this example, you will learn to find the GCD (Greatest Common Divisor) of two positive integers entered by the user using recursion.
This program takes two positive integers as input from the user and calculates
GCD using recursion.
Visit this page to learn how you can
calculate the GCD using loops.
GCD of Two Numbers using Recursion
#include <stdio.h>
int hcf(int n1, int n2);
int main() {
int n1, n2;
printf("Enter two positive integers: ");
scanf("%d %d", &n1, &n2);
printf("G.C.D of %d and %d is %d.", n1, n2, hcf(n1, n2));
return 0;
}
int hcf(int n1, int n2) {
if (n2 != 0)
return hcf(n2, n1 % n2);
else
return n1;
}

Output 
Enter two positive integers: 366
60
G.C.D of 366 and 60 is 6.
In this program, recursive calls are made until the value of n2 is
equal to 0.
Convert Binary Number to Decimal and vice-versa
In this example, you will learn to convert binary numbers to decimal and vice-versa manually by creating a user-defined function.
convert binary to decimal
#include <math.h>
#include <stdio.h>
int convert(long long n);
int main() {
long long n;
printf("Enter a binary number: ");
scanf("%lld", &n);
printf("%lld in binary = %d in decimal", n, convert(n));
return 0;
}
int convert(long long n) {
int dec = 0, i = 0, rem;
while (n != 0) {
rem = n % 10;
n /= 10;
dec += rem * pow(2, i);
++i;
}
return dec;
}

Output 
Enter a binary number: 110110111
110110111 in binary = 439
convert decimal to binary
#include <math.h>
#include <stdio.h>
long long convert(int n);
int main() {
int n;
printf("Enter a decimal number: ");
scanf("%d", &n);
printf("%d in decimal = %lld in binary", n, convert(n));
return 0;
}
long long convert(int n) {
long long bin = 0;
int rem, i = 1, step = 1;
while (n != 0) {
rem = n % 2;
printf("Step %d: %d/2, Remainder = %d, Quotient = %d\n", step++, n, rem, n / 2);
n /= 2;
bin += rem * i;
i *= 10;
}
return bin;
}
Output
Enter a decimal number: 19
Step 1: 19/2, Remainder = 1, Quotient = 9
Step 2: 9/2, Remainder = 1, Quotient = 4
Step 3: 4/2, Remainder = 0, Quotient = 2
Step 4: 2/2, Remainder = 0, Quotient = 1
Step 5: 1/2, Remainder = 1, Quotient = 0
19 in decimal = 10011 in binary
Convert Octal Number to Decimal and vice-versa
In this example, you will learn to convert octal numbers to decimal and  vice-versa manually by creating a user-defined function.
Program to Convert Decimal to Octal
#include <stdio.h>
#include <math.h>
int convertDecimalToOctal(int decimalNumber);
int main()
{
int decimalNumber;
printf("Enter a decimal number: ");
scanf("%d", &decimalNumber);
printf("%d in decimal = %d in octal", decimalNumber, convertDecimalToOctal(decimalNumber));
return 0;
}
int convertDecimalToOctal(int decimalNumber)
{
int octalNumber = 0, i = 1;
while (decimalNumber != 0)
{
octalNumber += (decimalNumber % 8) * i;
decimalNumber /= 8;
i *= 10;
}
return octalNumber;
}
Output
Enter a decimal number: 78
78 in decimal = 116 in octal
Program to Convert Octal to Decimal
#include <stdio.h>
#include <math.h>
long long convertOctalToDecimal(int octalNumber);
int main()
{
int octalNumber;
printf("Enter an octal number: ");
scanf("%d", &octalNumber);
printf("%d in octal = %lld in decimal", octalNumber, convertOctalToDecimal(octalNumber));
return 0;
}
long long convertOctalToDecimal(int octalNumber)
{
int decimalNumber = 0, i = 0;
while(octalNumber != 0)
{
decimalNumber += (octalNumber%10) * pow(8,i);
++i;
octalNumber/=10;
}
i = 1;
return decimalNumber;
}

Output
Enter an octal number: 116
116 in octal = 78 in decimal
Convert Binary Number to Octal and vice-versa
In this example, you will learn to convert binary numbers to octal and vice versa manually by creating a user-defined function.
Convert Binary to Octal
In this program, we will first convert a binary number to decimal. Then, the
decimal number is converted to octal.
#include <math.h>
#include <stdio.h>
int convert(long long bin);
int main() {
long long bin;
printf("Enter a binary number: ");
scanf("%lld", &bin);
printf("%lld in binary = %d in octal", bin, convert(bin));
return 0;
}
int convert(long long bin) {
int oct = 0, dec = 0, i = 0;
// converting binary to decimal
while (bin != 0) {
dec += (bin % 10) * pow(2, i);
++i;
bin /= 10;
}
i = 1;
// converting to decimal to octal
while (dec != 0) {
oct += (dec % 8) * i;
dec /= 8;
i *= 10;
}
return oct;
}
Output
Enter a binary number: 101001
101001 in binary = 51 in octal
Convert Octal to Binary
In this program, an octal number is converted to decimal at first. Then, the
decimal number is converted to binary number.
#include <math.h>
#include <stdio.h>
long long convert(int oct);
int main() {
int oct;
printf("Enter an octal number: ");
scanf("%d", &oct);
printf("%d in octal = %lld in binary", oct, convert(oct));
return 0;
}
long long convert(int oct) {
int dec = 0, i = 0;
long long bin = 0;
// converting octal to decimal
while (oct != 0) {
dec += (oct % 10) * pow(8, i);
++i;
oct /= 10;
}
i = 1;
// converting decimal to binary
while (dec != 0) {
bin += (dec % 2) * i;
dec /= 2;
i *= 10;
}
return bin;
}

Output 
Enter an octal number: 67
67 in octal = 110111 in binary
Reverse a Sentence Using Recursion
In this example, you will learn to take a sentence from the user and reverse it using recursion.
Reverse a sentence using recursion
#include <stdio.h>
void reverseSentence();
int main() {
printf("Enter a sentence: ");
reverseSentence();
return 0;
}
void reverseSentence() {
char c;
scanf("%c", &c);
if (c != '\n') {
reverseSentence();
printf("%c", c);
}
}

Output 
Enter a sentence: margorp emosewa
awesome program
This program first prints Enter a sentence: . Then, the reverseSentence() function is called.
This function stores the first letter entered by the user in c. If the variable is any character other than \n (newline), reverseSentence() is called again.
This process goes on until the user hits enter.
When the user hits enter, the reverseSentence() function starts printing characters from last.
calculate the power using recursion
In this example, you will learn to calculate the power of a number using recursion.
calculate power using recursion
#include <stdio.h>
int power(int n1, int n2);
int main() {
int base, a, result;
printf("Enter base number: ");
scanf("%d", &base);
printf("Enter power number(positive integer): ");
scanf("%d", &a);
result = power(base, a);
printf("%d^%d = %d", base, a, result);
return 0;
}
int power(int base, int a) {
if (a != 0)
return (base * power(base, a - 1));
else
return 1;
}

Output 
Enter base number: 3
Enter power number(positive integer): 4
3^4 = 81
You can also
compute the power of a number using a loop.
If you need to calculate the power of a number raised to a decimal value, you
can use the
pow() library function.
Calculate Average Using Arrays
In this example, you will learn to calculate the average of n number of elements entered by the user using arrays.
Store Numbers and Calculate Average Using Arrays
#include <stdio.h>
int main() {
int n, i;
float num[100], sum = 0.0, avg;
printf("Enter the numbers of elements: ");
scanf("%d", &n);
while (n > 100 || n < 1) {
printf("Error! number should in range of (1 to 100).\n");
printf("Enter the number again: ");
scanf("%d", &n);
}
for (i = 0; i < n; ++i) {
printf("%d. Enter number: ", i + 1);
scanf("%f", &num[i]);
sum += num[i];
}
avg = sum / n;
printf("Average = %.2f", avg);
return 0;
}

Output 
Enter the numbers of elements: 6
1. Enter number: 45.3
2. Enter number: 67.5
3. Enter number: -45.6
4. Enter number: 20.34
5. Enter number: 33
6. Enter number: 45.6
Average = 27.69
Here, the user is first asked to enter the number of elements. This number is
assigned to n.
If the user entered integer is greater less than 1 or greater than 100, the
user is asked to enter the number again. This is done using a
while loop.
Then, we have iterated a for loop from i = 0 to
i < n. In each iteration of the loop, the user is asked to enter
numbers to calculate the average. These numbers are stored in the
num[] array.
scanf("%f", &num[i]);
And, the sum of each entered element is computed.
sum += num[i];
Once the for loop is completed, the average is calculated and
printed on the screen.
  
Find Largest Element in an Array
In this example, you will learn to display the largest element entered by the user in an array.
Example: Largest Element in an array
#include <stdio.h>
int main() {
int n;
double arr[100];
printf("Enter the number of elements (1 to 100): ");
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
printf("Enter number%d: ", i + 1);
scanf("%lf", &arr[i]);
}
// storing the largest number to arr[0]
for (int i = 1; i < n; ++i) {
if (arr[0] < arr[i]) {
arr[0] = arr[i];
}
}
printf("Largest element = %.2lf", arr[0]);
return 0;
}
Output
Enter the number of elements (1 to 100): 5
Enter number1: 34.5
Enter number2: 2.4
Enter number3: -35.5
Enter number4: 38.7
Enter number5: 24.5
Largest element = 38.70
This program takes n number of elements from the user and stores it in the arr array.
To find the largest element,
the first two elements of array are checked and the largest of these two elements are placed in arr[0]
the first and third elements are checked and largest of these two elements is placed in arr[0].
this process continues until the first and last elements are checked
the largest number will be stored in the arr[0] position
// storing the largest number at arr[0]
for (int i = 1; i & lt; n; ++i) {
if (arr[0] & lt; arr[i]) {
arr[0] = arr[i];
}
}  
Calculate Standard Deviation
In this example, you will learn to calculate the standard deviation of 10 numbers stored in an array.
This program calculates the standard deviation of an individual series using arrays. Visit this page to learn about Standard Deviation.
To calculate the standard deviation, we have created a function named calculateSD().
Example: Population Standard Deviation
// SD of a population
#include <math.h>
#include <stdio.h>
float calculateSD(float data[]);
int main() {
int i;
float data[10];
printf("Enter 10 elements: ");
for (i = 0; i < 10; ++i)
scanf("%f", &data[i]);
printf("\nStandard Deviation = %.6f", calculateSD(data));
return 0;
}
float calculateSD(float data[]) {
float sum = 0.0, mean, SD = 0.0;
int i;
for (i = 0; i < 10; ++i) {
sum += data[i];
}
mean = sum / 10;
for (i = 0; i < 10; ++i) {
SD += pow(data[i] - mean, 2);
}
return sqrt(SD / 10);
}

Output 
Enter 10 elements: 1
2
3
4
5
6
7
8
9
10
Standard Deviation = 2.872281
Here, the array containing 10 elements is passed to the calculateSD() function. The function calculates the standard deviation using mean and returns it.
Note: The program calculates the standard deviation of a population. If you need to find the standard deviation of a sample, the formula is slightly different.
Add Two Matrices Using Multi-dimensional Arrays
In this example, you will learn to add two matrices in C programming using two-dimensional arrays.
Add Two Matrices
#include <stdio.h>
int main() {
int r, c, a[100][100], b[100][100], sum[100][100], i, j;
printf("Enter the number of rows (between 1 and 100): ");
scanf("%d", &r);
printf("Enter the number of columns (between 1 and 100): ");
scanf("%d", &c);
printf("\nEnter elements of 1st matrix:\n");
for (i = 0; i < r; ++i)
for (j = 0; j < c; ++j) {
printf("Enter element a%d%d: ", i + 1, j + 1);
scanf("%d", &a[i][j]);
}
printf("Enter elements of 2nd matrix:\n");
for (i = 0; i < r; ++i)
for (j = 0; j < c; ++j) {
printf("Enter element b%d%d: ", i + 1, j + 1);
scanf("%d", &b[i][j]);
}
// adding two matrices
for (i = 0; i < r; ++i)
for (j = 0; j < c; ++j) {
sum[i][j] = a[i][j] + b[i][j];
}
// printing the result
printf("\nSum of two matrices: \n");
for (i = 0; i < r; ++i)
for (j = 0; j < c; ++j) {
printf("%d   ", sum[i][j]);
if (j == c - 1) {
printf("\n\n");
}
}
return 0;
}

Output 
Enter the number of rows (between 1 and 100): 2
Enter the number of columns (between 1 and 100): 3
Enter elements of 1st matrix:
Enter element a11: 2
Enter element a12: 3
Enter element a13: 4
Enter element a21: 5
Enter element a22: 2
Enter element a23: 3
Enter elements of 2nd matrix:
Enter element b11: -4
Enter element b12: 5
Enter element b13: 3
Enter element b21: 5
Enter element b22: 6
Enter element b23: 3
Sum of two matrices: 
-2   8   7   
10   8   6  
In this program, the user is asked to enter the number of rows r and columns c. Then, the user is asked to enter the elements of the two matrices (of order rxc).
We then added corresponding elements of two matrices and saved it in another matrix (two-dimensional array). Finally, the result is printed on the screen.
Multiply Two Matrices Using Multi-dimensional Arrays
In this example, you will learn to multiply two matrices and display it using user-defined functions.
This program asks the user to enter the size (rows and columns) of two matrices.
To multiply two matrices, the number of columns of the first matrix should be equal to the number of rows of the second matrix.
The program below asks for the number of rows and columns of two matrices until the above condition is satisfied.
Then, the multiplication of two matrices is performed, and the result is displayed on the screen.
To perform this, we have created three functions:
getMatrixElements() - to take matrix elements input from the user.
multiplyMatrices() - to multiply two matrices.
display() - to display the resultant matrix after multiplication.
Multiply Matrices by Passing it to a Function
#include <stdio.h>
// function to get matrix elements entered by the user
void getMatrixElements(int matrix[][10], int row, int column) {
printf("\nEnter elements: \n");
for (int i = 0; i < row; ++i) {
for (int j = 0; j < column; ++j) {
printf("Enter a%d%d: ", i + 1, j + 1);
scanf("%d", &matrix[i][j]);
}
}
}
// function to multiply two matrices
void multiplyMatrices(int first[][10],
int second[][10],
int result[][10],
int r1, int c1, int r2, int c2) {
// Initializing elements of matrix mult to 0.
for (int i = 0; i < r1; ++i) {
for (int j = 0; j < c2; ++j) {
result[i][j] = 0;
}
}
// Multiplying first and second matrices and storing it in result
for (int i = 0; i < r1; ++i) {
for (int j = 0; j < c2; ++j) {
for (int k = 0; k < c1; ++k) {
result[i][j] += first[i][k] * second[k][j];
}
}
}
}
// function to display the matrix
void display(int result[][10], int row, int column) {
printf("\nOutput Matrix:\n");
for (int i = 0; i < row; ++i) {
for (int j = 0; j < column; ++j) {
printf("%d  ", result[i][j]);
if (j == column - 1)
printf("\n");
}
}
}
int main() {
int first[10][10], second[10][10], result[10][10], r1, c1, r2, c2;
printf("Enter rows and column for the first matrix: ");
scanf("%d %d", &r1, &c1);
printf("Enter rows and column for the second matrix: ");
scanf("%d %d", &r2, &c2);
// Taking input until
// 1st matrix columns is not equal to 2nd matrix row
while (c1 != r2) {
printf("Error! Enter rows and columns again.\n");
printf("Enter rows and columns for the first matrix: ");
scanf("%d%d", &r1, &c1);
printf("Enter rows and columns for the second matrix: ");
scanf("%d%d", &r2, &c2);
}
// get elements of the first matrix
getMatrixElements(first, r1, c1);
// get elements of the second matrix
getMatrixElements(second, r2, c2);
// multiply two matrices.
multiplyMatrices(first, second, result, r1, c1, r2, c2);
// display the result
display(result, r1, c2);
return 0;
}

Output 
Enter rows and column for the first matrix: 2
3
Enter rows and column for the second matrix: 3
2
Enter elements:
Enter a11: 2  
Enter a12: -3
Enter a13: 4
Enter a21: 53
Enter a22: 3
Enter a23: 5
Enter elements:
Enter a11: 3
Enter a12: 3
Enter a21: 5
Enter a22: 0
Enter a31: -3
Enter a32: 4
Output Matrix:
-21  22
159  179
Find Transpose of a Matrix
In this example, you will learn to find the transpose of a matrix in C programming.
The transpose of a matrix is a new matrix that is obtained by exchanging the rows and columns.
In this program, the user is asked to enter the number of rows r and columns c. Their values should be less than 10 in this program.
Then, the user is asked to enter the elements of the matrix (of order r*c).
The program below then computes the transpose of the matrix and prints it on the screen.
Find the Transpose of a Matrix
#include <stdio.h>
int main() {
int a[10][10], transpose[10][10], r, c;
printf("Enter rows and columns: ");
scanf("%d %d", &r, &c);
// asssigning elements to the matrix
printf("\nEnter matrix elements:\n");
for (int i = 0; i < r; ++i)
for (int j = 0; j < c; ++j) {
printf("Enter element a%d%d: ", i + 1, j + 1);
scanf("%d", &a[i][j]);
}
// printing the matrix a[][]
printf("\nEntered matrix: \n");
for (int i = 0; i < r; ++i)
for (int j = 0; j < c; ++j) {
printf("%d  ", a[i][j]);
if (j == c - 1)
printf("\n");
}
// computing the transpose
for (int i = 0; i < r; ++i)
for (int j = 0; j < c; ++j) {
transpose[j][i] = a[i][j];
}
// printing the transpose
printf("\nTranspose of the matrix:\n");
for (int i = 0; i < c; ++i)
for (int j = 0; j < r; ++j) {
printf("%d  ", transpose[i][j]);
if (j == r - 1)
printf("\n");
}
return 0;
}

Output 
Enter rows and columns: 2
3
Enter matrix elements:
Enter element a11: 1
Enter element a12: 4
Enter element a13: 0
Enter element a21: -5
Enter element a22: 2
Enter element a23: 7
Entered matrix:
1  4  0
-5  2  7
Transpose of the matrix:
1  -5
4  2
0  7 
Multiply two Matrices by Passing Matrix to a Function
In this example, you'll learn to multiply two matrices and display it using user defined function.
This program asks the user to enter the size of the matrix (rows and column).
Then, it asks the user to enter the elements of those matrices and finally adds and displays the result.
To perform this task three functions are made:
To takes matrix elements from user enterData()
To multiply two matrix multiplyMatrices()
To display the resultant matrix after multiplication display()

Example: Multiply Matrices by Passing it to a Function
#include <stdio.h>
void enterData(int firstMatrix[][10], int secondMatrix[][10], int rowFirst, int columnFirst, int rowSecond, int columnSecond);
void multiplyMatrices(int firstMatrix[][10], int secondMatrix[][10], int multResult[][10], int rowFirst, int columnFirst, int rowSecond, int columnSecond);
void display(int mult[][10], int rowFirst, int columnSecond);
int main()
{
	int firstMatrix[10][10], secondMatrix[10][10], mult[10][10], rowFirst, columnFirst, rowSecond, columnSecond, i, j, k;
	printf("Enter rows and column for first matrix: ");
	scanf("%d %d", &rowFirst, &columnFirst);
	printf("Enter rows and column for second matrix: ");
	scanf("%d %d", &rowSecond, &columnSecond);
	// If colum of first matrix in not equal to row of second matrix, asking user to enter the size of matrix again.
	while (columnFirst != rowSecond)
	{
		printf("Error! column of first matrix not equal to row of second.\n");
		printf("Enter rows and column for first matrix: ");
		scanf("%d%d", &rowFirst, &columnFirst);
		printf("Enter rows and column for second matrix: ");
		scanf("%d%d", &rowSecond, &columnSecond);
	}
	// Function to take matrices data
enterData(firstMatrix, secondMatrix, rowFirst, columnFirst, rowSecond, columnSecond);
// Function to multiply two matrices.
multiplyMatrices(firstMatrix, secondMatrix, mult, rowFirst, columnFirst, rowSecond, columnSecond);
// Function to display resultant matrix after multiplication.
display(mult, rowFirst, columnSecond);
	return 0;
}
void enterData(int firstMatrix[][10], int secondMatrix[][10], int rowFirst, int columnFirst, int rowSecond, int columnSecond)
{
	int i, j;
	printf("\nEnter elements of matrix 1:\n");
	for(i = 0; i < rowFirst; ++i)
	{
		for(j = 0; j < columnFirst; ++j)
		{
			printf("Enter elements a%d%d: ", i + 1, j + 1);
			scanf("%d", &firstMatrix[i][j]);
		}
	}
	printf("\nEnter elements of matrix 2:\n");
	for(i = 0; i < rowSecond; ++i)
	{
		for(j = 0; j < columnSecond; ++j)
		{
			printf("Enter elements b%d%d: ", i + 1, j + 1);
			scanf("%d", &secondMatrix[i][j]);
		}
	}
}
void multiplyMatrices(int firstMatrix[][10], int secondMatrix[][10], int mult[][10], int rowFirst, int columnFirst, int rowSecond, int columnSecond)
{
	int i, j, k;
	// Initializing elements of matrix mult to 0.
	for(i = 0; i < rowFirst; ++i)
	{
		for(j = 0; j < columnSecond; ++j)
		{
			mult[i][j] = 0;
		}
	}
	// Multiplying matrix firstMatrix and secondMatrix and storing in array mult.
	for(i = 0; i < rowFirst; ++i)
	{
		for(j = 0; j < columnSecond; ++j)
		{
			for(k=0; k<columnFirst; ++k)
			{
				mult[i][j] += firstMatrix[i][k] * secondMatrix[k][j];
			}
		}
	}
}
void display(int mult[][10], int rowFirst, int columnSecond)
{
	int i, j;
	printf("\nOutput Matrix:\n");
	for(i = 0; i < rowFirst; ++i)
	{
		for(j = 0; j < columnSecond; ++j)
		{
			printf("%d  ", mult[i][j]);
			if(j == columnSecond - 1)
				printf("\n\n");
		}
	}
}
Output
Enter rows and column for first matrix: 3
2
Enter rows and column for second matrix: 3
2
Error! column of first matrix not equal to row of second.
Enter rows and column for first matrix: 2
3
Enter rows and column for second matrix: 3
2
Enter elements of matrix 1:
Enter elements a11: 3
Enter elements a12: -2
Enter elements a13: 5
Enter elements a21: 3
Enter elements a22: 0
Enter elements a23: 4
Enter elements of matrix 2:
Enter elements b11: 2
Enter elements b12: 3
Enter elements b21: -9
Enter elements b22: 0
Enter elements b31: 0
Enter elements b32: 4
Output Matrix:
24  29
6  25
Access Array Elements Using Pointer
In this example, you will learn to access elements of an array using a pointer.
Access Array Elements Using Pointers
#include <stdio.h>
int main() {
int data[5];
printf("Enter elements: ");
for (int i = 0; i < 5; ++i)
scanf("%d", data + i);
printf("You entered: \n");
for (int i = 0; i < 5; ++i)
printf("%d\n", *(data + i));
return 0;
}

Output 
Enter elements: 1
2
3
5
4
You entered: 
1
2
3
5
4
In this program, the elements are stored in the integer array data[].
Then, the elements of the array are accessed using the pointer notation. By the way,
data[0] is equivalent to *data and &data[0] is equivalent to data
data[1] is equivalent to *(data + 1) and &data[1] is equivalent to data + 1
data[2] is equivalent to *(data + 2) and &data[2] is equivalent to data + 2
...
data[i] is equivalent to *(data + i) and &data[i] is equivalent to data + i
Visit this page to learn about the relationship between pointers and arrays.
Swap Numbers in Cyclic Order Using Call by Reference
In this example, the three numbers entered by the user are swapped in cyclic order using call by reference.
Swap Elements Using Call by Reference
#include <stdio.h>
void cyclicSwap(int *a, int *b, int *c);
int main() {
int a, b, c;
printf("Enter a, b and c respectively: ");
scanf("%d %d %d", &a, &b, &c);
printf("Value before swapping:\n");
printf("a = %d \nb = %d \nc = %d\n", a, b, c);
cyclicSwap(&a, &b, &c);
printf("Value after swapping:\n");
printf("a = %d \nb = %d \nc = %d", a, b, c);
return 0;
}
void cyclicSwap(int *n1, int *n2, int *n3) {
int temp;
// swapping in cyclic order
temp = *n2;
*n2 = *n1;
*n1 = *n3;
*n3 = temp;
}

Output 
Enter a, b and c respectively: 1
2
3
Value before swapping:
a = 1 
b = 2 
c = 3
Value after swapping:
a = 3 
b = 1 
c = 2
Here, the three numbers entered by the user are stored in variables
a, b and c respectively. The addresses of
these numbers are passed to the cyclicSwap() function.
cyclicSwap(&a, &b, &c);
In the function definition of cyclicSwap(), we have assigned
these addresses to pointers.
cyclicSwap(int *n1, int *n2, int *n3) {
...
}
When n1, n2 and n3 inside
cyclicSwap() are changed, the values of a,
b and c inside main() are also changed.
Note: The cyclicSwap() function is not returning
anything.
  
Find Largest Number Using Dynamic Memory Allocation
In this example, you will learn to find the largest number entered by the user in a dynamically allocated memory.
Example: Find the Largest Element
#include <stdio.h>
#include <stdlib.h>
int main() {
int n;
double *data;
printf("Enter the total number of elements: ");
scanf("%d", &n);
// Allocating memory for n elements
data = (double *)calloc(n, sizeof(double));
if (data == NULL) {
printf("Error!!! memory not allocated.");
exit(0);
}
// Storing numbers entered by the user.
for (int i = 0; i < n; ++i) {
printf("Enter number%d: ", i + 1);
scanf("%lf", data + i);
}
// Finding the largest number
for (int i = 1; i < n; ++i) {
if (*data < *(data + i)) {
*data = *(data + i);
    }
}
printf("Largest number = %.2lf", *data);
free(data);
return 0;
}
Output
Enter the total number of elements: 5
Enter number1: 3.4
Enter number2: 2.4
Enter number3: -5
Enter number4: 24.2
Enter number5: 6.7
Largest number = 24.20
Explanation
In the program, we have asked the user to enter the total number of elements which is stored in the variable n. Then, we have allocated memory for n number of double values.
// Allocating memory for n double values
data = (double *)calloc(n, sizeof(double));
Then, we used a for loop to take n number of data from the user.
// Storing elements
for (int i = 0; i < n; ++i) {
printf("Enter Number%d: ", i + 1);
scanf("%lf", data + i);
}
Finally, we used another for loop to compute the largest number.
// Computing the largest number
for (int i = 1; i < n; ++i) {
if (*data < *(data + i))
*data = *(data + i);
}
}
Note: Instead of calloc(), it's also possible to solve this problem using the malloc() function.
Find the Frequency of Characters in a String
In this example, you will learn to find the frequency of a character in a string.
Find the Frequency of a Character
#include <stdio.h>
int main() {
char str[1000], ch;
int count = 0;
printf("Enter a string: ");
fgets(str, sizeof(str), stdin);
printf("Enter a character to find its frequency: ");
scanf("%c", &ch);
for (int i = 0; str[i] != '\0'; ++i) {
if (ch == str[i])
++count;
}
printf("Frequency of %c = %d", ch, count);
return 0;
}

Output 
Enter a string: This website is awesome.
Enter a character to find its frequency: e
Frequency of e = 4
In this program, the string entered by the user is stored in str.
Then, the user is asked to enter the character whose frequency is to be found. This is stored in variable ch.
Then, a for loop is used to iterate over characters of the string. In each iteration, if the character in the string is equal to the ch, count is increased by 1.
Finally, the frequency stored in the count variable is printed.
Count the Number of Vowels, Consonants and so on
In this example, the number of vowels, consonants, digits, and white-spaces in a string entered by the user is counted.
count vowels, consonants etc.
#include <stdio.h>
int main() {
char line[150];
int vowels, consonant, digit, space;
vowels = consonant = digit = space = 0;
printf("Enter a line of string: ");
fgets(line, sizeof(line), stdin);
for (int i = 0; line[i] != '\0'; ++i) {
if (line[i] == 'a' || line[i] == 'e' || line[i] == 'i' ||
line[i] == 'o' || line[i] == 'u' || line[i] == 'A' ||
line[i] == 'E' || line[i] == 'I' || line[i] == 'O' ||
line[i] == 'U') {
++vowels;
} else if ((line[i] >= 'a' && line[i] <= 'z') || (line[i] >= 'A' && line[i] <= 'Z')) {
++consonant;
} else if (line[i] >= '0' && line[i] <= '9') {
++digit;
} else if (line[i] == ' ') {
++space;
}
}
printf("Vowels: %d", vowels);
printf("\nConsonants: %d", consonant);
printf("\nDigits: %d", digit);
printf("\nWhite spaces: %d", space);
return 0;
}

Output 
Enter a line of string: adfslkj34 34lkj343 34lk
Vowels: 1
Consonants: 11
Digits: 9
White spaces: 2
Here, the string entered by the user is stored in the
line variable.
Initially, the variables vowel, consonant,
digit, and space are initialized to 0.
Then, a for loop is used to iterate over characters of a string.
In each iteration, whether the character is vowel, consonant, digit, and space
is checked. Suppose, the character is a vowel, in this case, the
vowel variable is increased by 1.
When the loop ends, the number of vowels, consonants, digits and white spaces
are stored in variables vowel, consonant,
digit and space respectively.
Remove all Characters in a String Except Alphabets
In this example, you will learn to remove all the characters from a string entered by the user except the alphabets.
Remove Characters in String Except Alphabets
#include <stdio.h>
int main() {
char line[150];
printf("Enter a string: ");
fgets(line, sizeof(line), stdin); // take input
for (int i = 0, j; line[i] != '\0'; ++i) {
// enter the loop if the character is not an alphabet
// and not the null character
while (!(line[i] >= 'a' && line[i] <= 'z') && !(line[i] >= 'A' && line[i] <= 'Z') && !(line[i] == '\0')) {
for (j = i; line[j] != '\0'; ++j) {
// if jth element of line is not an alphabet,
// assign the value of (j+1)th element to the jth element
line[j] = line[j + 1];
}
line[j] = '\0';
}
}
printf("Output String: ");
puts(line);
return 0;
}

Output 
Enter a string: p2'r-o@gram84iz./
Output String: programiz
This program takes a string input from the user and stores in the line variable. Then, a for loop is used to iterate over characters of the string.
If the character in a string is not an alphabet, it is removed from the string and the position of the remaining characters are shifted to the left by 1 position.
Find the Length of a String
In this example, you will learn to find the length of a string manually without using the strlen() function.
As you know, the best way to find the length of a string is by using the
strlen() function. However, in this example, we will find the
length of a string manually.
Calculate Length of String without Using strlen() Function
#include <stdio.h>
int main() {
char s[] = "Programming is fun";
int i;
for (i = 0; s[i] != '\0'; ++i);
printf("Length of the string: %d", i);
return 0;
}

Output 
Length of the string: 18
Here, using a for loop, we have iterated over characters of the
string from i = 0 to until '\0' (null character) is
encountered. In each iteration, the value of i is increased by 1.
When the loop ends, the length of the string will be stored in the
i variable.
  
Concatenate Two Strings
In this example, you will learn to concatenate two strings manually without using the strcat() function.
As you know, the best way to concatenate two strings in C programming is by using the strcat() function. However, in this example, we will concatenate two strings manually.
Concatenate Two Strings Without Using strcat()
#include <stdio.h>
int main() {
char s1[100] = "programming ", s2[] = "is awesome";
int length, j;
// store length of s1 in the length variable
length = 0;
while (s1[length] != '\0') {
++length;
}
// concatenate s2 to s1
for (j = 0; s2[j] != '\0'; ++j, ++length) {
s1[length] = s2[j];
}
// terminating the s1 string
s1[length] = '\0';
printf("After concatenation: ");
puts(s1);
return 0;
}

Output 
After concatenation: programming is awesome
Here, two strings s1 and s2 and concatenated and the result is stored in s1.
It's important to note that the length of s1 should be sufficient to hold the string after concatenation. If not, you may get unexpected output.
Copy String Without Using strcpy()
In this example, you will learn to copy strings without using the strcpy() function.
As you know, the best way to copy a string is by using the
strcpy() function. However, in this example, we will copy a
string manually without using the strcpy() function.
Copy String Without Using strcpy()
#include <stdio.h>
int main() {
char s1[100], s2[100], i;
printf("Enter string s1: ");
fgets(s1, sizeof(s1), stdin);
for (i = 0; s1[i] != '\0'; ++i) {
s2[i] = s1[i];
}
s2[i] = '\0';
printf("String s2: %s", s2);
return 0;
}

Output 
Enter string s1: Hey fellow programmer.
String s2: Hey fellow programmer.
The above program copies the content of string s1 to string
s2 manually.
Sort Elements in Lexicographical Order (Dictionary Order)
In this example, you will learn to sort 5 strings entered by the user in the lexicographical order (dictionary order).
Sort strings in the dictionary order
#include <stdio.h>
#include <string.h>
int main() {
char str[5][50], temp[50];
printf("Enter 5 words: ");
// Getting strings input
for (int i = 0; i < 5; ++i) {
fgets(str[i], sizeof(str[i]), stdin);
}
// storing strings in the lexicographical order
for (int i = 0; i < 5; ++i) {
for (int j = i + 1; j < 5; ++j) {
// swapping strings if they are not in the lexicographical order
if (strcmp(str[i], str[j]) > 0) {
strcpy(temp, str[i]);
strcpy(str[i], str[j]);
strcpy(str[j], temp);
}
}
}
printf("\nIn the lexicographical order: \n");
for (int i = 0; i < 5; ++i) {
fputs(str[i], stdout);
}
return 0;
}
Output
Enter 5 words: R programming
JavaScript
Java
C programming
C++ programming
In the lexicographical order:
C programming
C++ programming
Java
JavaScript
R programming
To solve this program, a two-dimensional string named str is created. The string can hold a maximum of 5 strings and each string can have a maximum of 50 characters (including the null character).
In the program, we have used two library functions:
strcmp() - to compare strings
strcpy() - to copy strings
These functions are used to compare strings and sort them in the correct order.
Store Information of a Student Using Structure
In this example, you will learn to store the information of a student in a structure and display them on the screen.
Store Information and Display it Using Structure
#include <stdio.h>
struct student {
char name[50];
int roll;
float marks;
} s;
int main() {
printf("Enter information:\n");
printf("Enter name: ");
fgets(s.name, sizeof(s.name), stdin);
printf("Enter roll number: ");
scanf("%d", &s.roll);
printf("Enter marks: ");
scanf("%f", &s.marks);
printf("Displaying Information:\n");
printf("Name: ");
printf("%s", s.name);
printf("Roll number: %d\n", s.roll);
printf("Marks: %.1f\n", s.marks);
return 0;
}

Output 
Enter information:
Enter name: Jack
Enter roll number: 23
Enter marks: 34.5
Displaying Information:
Name: Jack
Roll number: 23
Marks: 34.5
In this program, a structure student is created. The structure has
three members: name (string), roll (integer) and
marks (float).
Then, a structure variable s is created to store information and
display it on the screen.
  
Add Two Distances (in inch-feet system) using Structures
In this example, you will learn to take two distances (in the inch-feet system), add them and display the result on the screen.
If you do not know, 12 inches is 1 foot.
add two distances in the inch-feet system
#include <stdio.h>
struct Distance {
int feet;
float inch;
} d1, d2, result;
int main() {
// take first distance input
printf("Enter 1st distance\n");
printf("Enter feet: ");
scanf("%d", &d1.feet);
printf("Enter inch: ");
scanf("%f", &d1.inch);
// take second distance input
printf("\nEnter 2nd distance\n");
printf("Enter feet: ");
scanf("%d", &d2.feet);
printf("Enter inch: ");
scanf("%f", &d2.inch);
// adding distances
result.feet = d1.feet + d2.feet;
result.inch = d1.inch + d2.inch;
// convert inches to feet if greater than 12
while (result.inch >= 12.0) {
result.inch = result.inch - 12.0;
++result.feet;
}
printf("\nSum of distances = %d\'-%.1f\", result.feet, result.inch);
return 0;
}

Output 
Enter 1st distance
Enter feet: 23
Enter inch: 8.6
Enter 2nd distance
Enter feet: 34
Enter inch: 2.4
Sum of distances = 57'-11.0"
In this program, a structure Distance is defined. The structure has two members:
feet - an integer
inch - a float
Two variables d1 and d2 of type struct Distance are created. These variables store distances in the feet and inches.
Then, the sum of these two distances are computed and stored in the result variable. Finally, result is printed on the screen.
Add Two Complex Numbers by Passing Structure to a Function
In this example, you will learn to take two complex numbers as structures and add them by creating a user-defined function.
Add Two Complex Numbers
#include <stdio.h>
typedef struct complex {
float real;
float imag;
} complex;
complex add(complex n1, complex n2);
int main() {
complex n1, n2, result;
printf("For 1st complex number \n");
printf("Enter the real and imaginary parts: ");
scanf("%f %f", &n1.real, &n1.imag);
printf("\nFor 2nd complex number \n");
printf("Enter the real and imaginary parts: ");
scanf("%f %f", &n2.real, &n2.imag);
result = add(n1, n2);
printf("Sum = %.1f + %.1fi", result.real, result.imag);
return 0;
}
complex add(complex n1, complex n2) {
complex temp;
temp.real = n1.real + n2.real;
temp.imag = n1.imag + n2.imag;
return (temp);
}

Output 
For 1st complex number
Enter the real and imaginary parts: 2.1
-2.3
For 2nd complex number
Enter the real and imaginary parts: 5.6
23.2
Sum = 7.7 + 20.9i
In this program, a structure named complex is declared. It has
two members: real and imag. We then created two
variables n1 and n2 from this structure.
These two structure variables are passed to the add() function.
The function computes the sum and returns the structure containing the sum.
Finally, the sum of complex numbers is printed from the
main() function.
Calculate Difference Between Two Time Periods
In this example, you will learn to calculate the difference between two time periods using a user-defined function.
Calculate Difference Between Two Time Periods
#include <stdio.h>
struct TIME {
int seconds;
int minutes;
int hours;
};
void differenceBetweenTimePeriod(struct TIME t1,
struct TIME t2,
struct TIME *diff);
int main() {
struct TIME startTime, stopTime, diff;
printf("Enter the start time. \n");
printf("Enter hours, minutes and seconds: ");
scanf("%d %d %d", &startTime.hours,
&startTime.minutes,
&startTime.seconds);
printf("Enter the stop time. \n");
printf("Enter hours, minutes and seconds: ");
scanf("%d %d %d", &stopTime.hours,
&stopTime.minutes,
&stopTime.seconds);
// Difference between start and stop time
differenceBetweenTimePeriod(startTime, stopTime, &diff);
printf("\nTime Difference: %d:%d:%d - ", startTime.hours,
startTime.minutes,
startTime.seconds);
printf("%d:%d:%d ", stopTime.hours,
stopTime.minutes,
stopTime.seconds);
printf("= %d:%d:%d\n", diff.hours,
diff.minutes,
diff.seconds);
return 0;
}
// Computes difference between time periods
void differenceBetweenTimePeriod(struct TIME start,
struct TIME stop,
struct TIME *diff) {
while (stop.seconds > start.seconds) {
--start.minutes;
start.seconds += 60;
}
diff->seconds = start.seconds - stop.seconds;
while (stop.minutes > start.minutes) {
--start.hours;
start.minutes += 60;
}
diff->minutes = start.minutes - stop.minutes;
diff->hours = start.hours - stop.hours;
}

Output 
Enter the start time.
Enter hours, minutes and seconds: 13
34
55
Enter the stop time.
Enter hours, minutes and seconds: 8
12
15
Time Difference: 13:34:55 - 8:12:15 = 5:22:40
In this program, the user is asked to enter two time periods and these two periods are stored in structure variables startTime and stopTime respectively.
Then, the function differenceBetweenTimePeriod() calculates the difference between the time periods. The result is displayed from the main() function without returning it (using call by reference technique).
Store Information of Students Using Structure
In this example, you will learn to store the information of 5 students by using an array of structures.
Store Information in Structure and Display it
#include <stdio.h>
struct student {
char firstName[50];
int roll;
float marks;
} s[10];
int main() {
int i;
printf("Enter information of students:\n");
// storing information
for (i = 0; i < 5; ++i) {
s[i].roll = i + 1;
printf("\nFor roll number%d,\n", s[i].roll);
printf("Enter first name: ");
scanf("%s", s[i].firstName);
printf("Enter marks: ");
scanf("%f", &s[i].marks);
}
printf("Displaying Information:\n\n");
// displaying information
for (i = 0; i < 5; ++i) {
printf("\nRoll number: %d\n", i + 1);
printf("First name: ");
puts(s[i].firstName);
printf("Marks: %.1f", s[i].marks);
printf("\n");
}
return 0;
}

Output 
Enter information of students: 
For roll number1,
Enter name: Tom
Enter marks: 98
For roll number2,
Enter name: Jerry
Enter marks: 89
.
.
.
Displaying Information:
Roll number: 1
Name: Tom
Marks: 98
.
.
.
In this program, a structure student is created. The structure
has three members: name (string), roll (integer) and
marks (float).
Then, we created an array of structures s having 5 elements to
store information of 5 students.
Using a for loop, the program takes the information of 5 students
from the user and stores it in the array of structure. Then using another
for loop, the information entered by the user is displayed on the
screen.
Store Data in Structures Dynamically
In this example, you will learn to store the information entered by the user using dynamic memory allocation.
This program asks the user to store the value of noOfRecords and allocates the memory for the noOfRecords structure variables dynamically using the malloc() function.
Demonstrate the Dynamic Memory Allocation for Structure
#include <stdio.h>
#include <stdlib.h>
struct course {
int marks;
char subject[30];
};
int main() {
struct course *ptr;
int noOfRecords;
printf("Enter the number of records: ");
scanf("%d", &noOfRecords);
// Memory allocation for noOfRecords structures
ptr = (struct course *)malloc(noOfRecords * sizeof(struct course));
for (int i = 0; i < noOfRecords; ++i) {
printf("Enter subject and marks:\n");
scanf("%s %d", (ptr + i)->subject, &(ptr + i)->marks);
}
printf("Displaying Information:\n");
for (int i = 0; i < noOfRecords; ++i) {
printf("%s\t%d\n", (ptr + i)->subject, (ptr + i)->marks);
}
free(ptr);
return 0;
}

Output 
Enter the number of records: 2
Enter subject and marks:
Science 82
Enter subject and marks:
DSA 73
Displaying Information:
Science     82
DSA     73
Write a Sentence to a File
In this example, you will learn to write a sentence in a file using fprintf() statement.
This program stores a sentence entered by the user in a file.
#include <stdio.h>
#include <stdlib.h>
int main() {
char sentence[1000];
// creating file pointer to work with files
FILE *fptr;
// opening file in writing mode
fptr = fopen("program.txt", "w");
// exiting program 
if (fptr == NULL) {
printf("Error!");
exit(1);
}
printf("Enter a sentence:\n");
fgets(sentence, sizeof(sentence), stdin);
fprintf(fptr, "%s", sentence);
fclose(fptr);
return 0;
}

Output 
Enter a sentence: C Programming is fun
Here, a file named program.txt is created. The file will contain C programming is fun text.
In the program, the sentence entered by the user is stored in the
sentence variable.
Then, a file named program.txt is opened in writing mode. If
the file does not exist, it will be created.
Finally, the string entered by the user will be written to this file using the
fprintf() function and the file is closed.
Read the First Line From a File
In this example, you will learn to print the first line from a file in C programming.
read the first line from a file
#include <stdio.h>
#include <stdlib.h> // For exit() function
int main() {
char c[1000];
FILE *fptr;
if ((fptr = fopen("program.txt", "r")) == NULL) {
printf("Error! File cannot be opened.");
// Program exits if the file pointer returns NULL.
exit(1);
}
// reads text until newline is encountered
fscanf(fptr, "%[^\n]", c);
printf("Data from the file:\n%s", c);
fclose(fptr);
return 0;
}
If the file is found, the program saves the content of the file to a string c until '\n' newline is encountered.
Suppose the program.txt file contains the following text in the current directory.
C programming is awesome.
I love C programming.
How are you doing? 
The output of the program will be:
Data from the file:
C programming is awesome.
If the file program.txt is not found, the program prints the error message.
Display its own Source Code as Output
In this example, you'll learn to display source of the program using __FILE__ macro.
Though this problem seems complex, the concept behind this program is straightforward; display the content from the same file you are writing the source code.

In C programming, there is a predefined macro named __FILE__ that gives the name of the current input file.
#include <stdio.h>
int main() {
   // location the current input file.
printf("%s",__FILE__);
}
C program to display its own source code
#include <stdio.h>
int main() {
FILE *fp;
int c;
   
    // open the current input file
fp = fopen(__FILE__,"r");
do {
c = getc(fp);   // read character 
putchar(c);     // display character
}
while(c != EOF);  // loop until the end of file is reached
    fclose(fp);
return 0;
}
Print Pyramids and Patterns
In this example, you will learn to print half pyramids, inverted pyramids, full pyramids, inverted full pyramids, Pascal's triangle, and Floyd's triangle in C Programming.
Example 1: Half Pyramid of *
*
* *
* * *
* * * *
* * * * *
C Program
#include <stdio.h>
int main() {
int i, j, rows;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 1; i <= rows; ++i) {
for (j = 1; j <= i; ++j) {
printf("* ");
}
printf("\n");
}
return 0;
}
Example 2: Half Pyramid of Numbers
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
C Program
#include <stdio.h>
int main() {
int i, j, rows;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 1; i <= rows; ++i) {
for (j = 1; j <= i; ++j) {
printf("%d ", j);
}
printf("\n");
}
return 0;
}
Example 3: Half Pyramid of Alphabets
A
B B
C C C
D D D D
E E E E E
C Program
#include <stdio.h>
int main() {
int i, j;
char input, alphabet = 'A';
printf("Enter an uppercase character you want to print in the last row: ");
scanf("%c", &input);
for (i = 1; i <= (input - 'A' + 1); ++i) {
for (j = 1; j <= i; ++j) {
printf("%c ", alphabet);
}
++alphabet;
printf("\n");
}
return 0;
}
Example 4: Inverted half pyramid of *
* * * * *
* * * *
* * * 
* *
*
C Program
#include <stdio.h>
int main() {
int i, j, rows;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = rows; i >= 1; --i) {
for (j = 1; j <= i; ++j) {
printf("* ");
}
printf("\n");
}
return 0;
}
Example 5: Inverted half pyramid of numbers
1 2 3 4 5
1 2 3 4 
1 2 3
1 2
1
C Program
#include <stdio.h>
int main() {
int i, j, rows;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = rows; i >= 1; --i) {
for (j = 1; j <= i; ++j) {
printf("%d ", j);
}
printf("\n");
}
return 0;
}
Example 6: Full Pyramid of *
        *
* * *
* * * * *
* * * * * * *
* * * * * * * * *
C Program
#include <stdio.h>
int main() {
int i, space, rows, k = 0;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 1; i <= rows; ++i, k = 0) {
for (space = 1; space <= rows - i; ++space) {
printf("  ");
}
while (k != 2 * i - 1) {
printf("* ");
++k;
}
printf("\n");
}
return 0;
}
Example 7: Full Pyramid of Numbers
        1
2 3 2
3 4 5 4 3
4 5 6 7 6 5 4
5 6 7 8 9 8 7 6 5
C Program
#include <stdio.h>
int main() {
int i, space, rows, k = 0, count = 0, count1 = 0;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 1; i <= rows; ++i) {
for (space = 1; space <= rows - i; ++space) {
printf("  ");
++count;
}
while (k != 2 * i - 1) {
if (count <= rows - 1) {
printf("%d ", i + k);
++count;
} else {
++count1;
printf("%d ", (i + k - 2 * count1));
}
++k;
}
count1 = count = k = 0;
printf("\n");
}
return 0;
}
Example 8: Inverted full pyramid of *
* * * * * * * * *
* * * * * * *
* * * * *
* * *
*
C Program
#include <stdio.h>
int main() {
int rows, i, j, space;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = rows; i >= 1; --i) {
for (space = 0; space < rows - i; ++space)
printf("  ");
for (j = i; j <= 2 * i - 1; ++j)
printf("* ");
for (j = 0; j < i - 1; ++j)
printf("* ");
printf("\n");
}
return 0;
}
Example 9: Pascal's Triangle
           1
1   1
1   2   1
1   3   3    1
1  4    6   4   1
1  5   10   10  5   1
C Program
#include <stdio.h>
int main() {
int rows, coef = 1, space, i, j;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 0; i < rows; i++) {
for (space = 1; space <= rows - i; space++)
printf("  ");
for (j = 0; j <= i; j++) {
if (j == 0 || i == 0)
coef = 1;
else
coef = coef * (i - j + 1) / j;
printf("%4d", coef);
}
printf("\n");
}
return 0;
}
Example 10: Floyd's Triangle.
1
2 3
4 5 6
7 8 9 10
C Program
#include <stdio.h>
int main() {
int rows, i, j, number = 1;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 1; i <= rows; i++) {
for (j = 1; j <= i; ++j) {
printf("%d ", number);
++number;
}
printf("\n");
}
return 0;
}