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Floating Point to Binary Value(C++)

Asked 2009-01-23T18:45:28.580
21

I want to take a floating point number in C++, like 2.25125, and a int array filled with the binary value that is used to store the float in memory (IEEE 754).

So I could take a number, and end up with a int num[16] array with the binary value of the float: num[0] would be 1 num[1] would be 1 num[2] would be 0 num[3] would be 1 and so on...

Putting an int into an array isn't difficult, just the process of getting the binary value of a float is where I'm stuck. Can you just read the binary in the memory that the float variable? If not, how could I go about doing this in C++?

EDIT: The reason for doing the comparison this way is that I am wanting to learn to do bitwise operations in C++.

c++ binary floating-point ieee-754
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2 Answers

15
int fl = *(int*)&floatVar; //assuming sizeof(int) = sizeof(float)

int binaryRepresentation[sizeof(float) * 8];

for (int i = 0; i < sizeof(float) * 8; ++i)
    binaryRepresentation[i] = ((1 << i) & fl) != 0 ? 1 : 0;

Explanation

(1 << i) shifts the value 1, i bits to the left. The & operator computes the bitwise and of the operands.

The for loop runs once for each of the 32 bits in the float. Each time, i will be the number of the bit we want to extract the value from. We compute the bitwise and of the number and 1 << i:

Assume the number is: 1001011, and i = 2

1<<i will be equal to 0000100

  10001011
& 00000100
==========
  00000000

if i = 3 then:

  10001011
& 00001000
==========
  00001000

Basically, the result will be a number with ith bit set to the ith bit of the original number and all other bits are zero. The result will be either zero, which means the ith bit in the original number was zero or nonzero, which means the actual number had the ith bit equal to 1.

answered 2009-01-23T18:47:55.673
1

Create a union of float and and unsigned long. set the value of the float member and iterate over the bits of the unsigned long value as already described in other answers.

This will eliminate the cast operators.

answered 2009-01-23T18:55:25.237

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