Consider below example

Convert 2.625 to binary.

We will consider the integer and fractional part separately.

The integral part is easy, 2 = 10. 

For the fractional part:

0.625 × 2 = 1.25 1 Generate 1 and continue with the rest. 0.25 × 2 = 0.5 0 Generate 0 and continue. 0.5 × 2 = 1.0 1 Generate 1 and nothing remains. 

So 0.625 = 0.101, and 2.625 = 10.101.

See this link for more information.

(d means decimal, b means binary)

1. 12.25d is your float.
2. You write 12d in binary and remove it from your float. Only the remainder (.25d) will be left.
3. You write the dot.
4. While the remainder (0.25d) is not zero (and/or you want more digits), multiply it with 2 (-> 0.50d), remove and write the digit left of the dot (0), and continue with the new remainder (.50d).

The float value is stored in IEEE 754 format so we can't convert it directly like integer, char to binary.

But we can convert float to binary through a pointer.

#include <stdio.h> int main() { float a = 7.5; int i; int * p; p = &a; for (i = sizeof(int) * 8 - 1; i >= 0; i--) { printf("%d", (*p) >> i & 1); } return 0; } 

Output

0 10000001 11100000000000000000000 

Spaces added for clarification, they are not included as part of the program.

x = float(raw_input("enter number between 0 and 1: ")) p = 0 while ((2**p)*x) %1 != 0: p += 1 # print p num = int (x * (2 ** p)) # print num result = '' if num == 0: result = '0' while num > 0: result = str(num%2) + result num = num / 2 for i in range (p - len(result)): result = '0' + result result = result[0:-p] + '.' + result[-p:] print result #this will print result for the decimal portion 

void transfer(double x) { unsigned long long * p = (unsigned long long * ) & x; for (int i = sizeof(unsigned long long) * 8 - 1; i >= 0; i--) { cout << (( * p) >> i & 1); } }