# C Program to convert a decimal number to a binary number

Last updated on July 27, 2020

The following is a C program to convert a decimal number to a binary number:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 /******************************************************* Program to convert a decimal number to a binary number ******************************************************/ #include // include stdio.h library #include // include math.h library int main(void) { long long num, bin = 0; int i = 0, rem; printf("Enter a decimal number: "); scanf("%lld", &num); while(num != 0) { rem = num % 2; // get the remainder bin = rem * (long long)pow(10, i++) + bin; num /= 2; // get the quotient } printf("%lld", bin); return 0; // return 0 to operating system } 

Expected Output: 1sr run:

 1 2 Enter a decimal number: 4 100 

2nd run:

 1 2 Enter a decimal number: 123456 11110001001000000 

## How it works: #

To convert a decimal number to a binary number, we follow these steps:

Step 1: Divide the decimal number continuously by 2 and right the remainder on the right-hand side of the dividend. We repeat this process until we get the quotient 0.

Step 2: Write the remainders from bottom to top.

Let's take some examples now.

Example 1: Convert decimal number 5 to a binary.

Step 1:

Quotient Remainder
5/2 2 1
2/2 1 0
1/2 0 1

Step 2:

$$5_{10}$$ = $$101_2$$

Example 2: Convert decimal number 123 to a binary.

Step 1:

Quotient Remainder
123/2 61 1
61/2 30 1
30/2 15 0
15/2 7 1
7/2 3 1
3/2 1 1
1/2 0 1

Step 2: $$123_{10}$$ = $$1111011_2$$

The following table shows what happens at each iteration of the loop (assuming num = 4):

Iteration rem bin num i
After 1st iteration rem = 4 % 2 = 0 bin = 0 * (10^0) + 0 = 0 num = 4 / 2 = 2 i = 2
After 2nd iteration rem = 2 % 2 = 0 bin = 0 * (10^1) + 0 = 0 num = 2 / 2 = 1 i = 3
After 3rd iteration rem = 1 % 2 = 1 bin = 1 * (10^2) + 0 = 100 num = 1 / 2 = 0 i = 4