Print bits of float number does not work, can you explain me why?

Question

I want to print the bits of a float number.

My plan was to:

Discover how many bits there are at all
Create and int array and insert into it the bits of the number using simple right bits shifting.
Because the bits were inserted in a reversed order to what I need, lets run another loop to go over the array one again but this time, from the last element to the first one.

Why?

Lets take for example the number 4.2: I believe the bits were entered in the following way:

4 - 1000

2 - 0010

So together it's 10000010.

Enters FILO - First in -> Last out. So 0 will be the first element but as you cans see here, we need it in the end.

Here is my code:

float FloatAnalysis(float number)
{
    int arr[32] = {0};
    float num_cpy = number;
    size_t num_of_bits = 0, i = 0;
    
    while (0 != number)
    {
        num_cpy >>= 1;
        ++num_of_bits;
    }
    
    num_cpy = number;
    
    for(i = 0; i < num_of_bits; ++i)
    {
    
        arr[i] = (num_cpy & 1);
        num_cpy >>= 1;
    
    }
    
    
    for(i = num_of_bits-1; i => 0; --i)
    {
    
        printf("%d", arr[i]);
    
    }
    
}

And here the output:

bitwise.c:359:11: error: invalid operands to binary >> (have ‘float’ and ‘int’)
  359 |   num_cpy >>= 1;
      |           ^~~
bitwise.c:368:21: error: invalid operands to binary & (have ‘float’ and ‘int’)
  368 |   arr[i] = (num_cpy & 1);
      |                     ^
bitwise.c:369:11: error: invalid operands to binary >> (have ‘float’ and ‘int’)
  369 |   num_cpy >>= 1;
      |           ^~~

Can you expl

ain me what is going on here?

`memcpy` the `float` into an `unsigned int` and work with that. — dbush, Mar 02 '21 at 03:53
Just like it says, you can't use `>>` and `&` on `float`; if you read the C standard or any book you'll see they're only defined for integer types. You have to get those bits into an integer first, using `memcpy` as dbush says, or a union. — Nate Eldredge, Mar 02 '21 at 04:25
Also note, floating point values are stored in IEEE-754 floating point format. For a 32-bit `float` that is 1-sign bit, 8-bits for the biased exponent and a 23-bit mantissa. You won't find your example of `4.2` as you expect. — David C. Rankin, Mar 02 '21 at 05:43
4.2 as a `float` is actually `0x40866666` which is actually 4.19999980926513671875. In other words, in IEEE-754 floating point, there is no such number as 4.2. — Steve Summit, Mar 02 '21 at 14:07

score 1 · Answer 1 · answered Mar 02 '21 at 14:01

Use memcpy

You cannot perform bitwise operations on a float.

You can use memcpy to copy your float to an unsigned int and preserves its bits:

float num_cpy = number;

becomes

unsigned int num_cpy;
memcpy(&num_cpy, &number, sizeof(unsigned));

Note that if you try to cast the result, by taking your float address in memory and cast it as unsigned, with:

num_cpy = *(float *)&number;

You will strip the floating point part away, you will preserve the value (or what can be preserved) but loose the accuracy of its binary representation.

Example

In the below example,

float number = 42.42;
unsigned int num_cpy;
memcpy(&num_cpy, &number, sizeof(unsigned)); 
unsigned int num_cpy2 = *(float *)&number;
printf("Bits in num_cpy: %d    bits in num_cpy2: %d\n", __builtin_popcount(num_cpy), __builtin_popcount(num_cpy2));
printf("%d\n", num_cpy);
printf("%d\n", num_cpy2);

will output

Bits in num_cpy: 12    bits in num_cpy2: 3
1110027796 // memcpy
42 // cast

More reading

I recommend that you especially take a look at floating point internal representation that sums up very well what is going at the bits level.

Internal Representation: sign: 1 bit, exponent: 8 bits, fraction: 23 bits
(for a single precision, 32 bits floating point, that we call float in C)

Given the `float number;` definition, `*(float *)&number` is meaningless. And something like `*(unsigned int *)&number` is a strict aliasing violation and therefore undefined behavior that [can fail even on x86 systems](https://stackoverflow.com/questions/47510783/why-does-unaligned-access-to-mmaped-memory-sometimes-segfault-on-amd64). — Andrew Henle, Mar 02 '21 at 14:15

score 1 · Answer 2 · answered Mar 23 '21 at 23:04

OP's code has various problems aside from compiler error.

i => 0 is not proprer code. Perhaps OP wanted i >= 0?. Even that has trouble.

size_t num_of_bits = 0, i = 0;
...
//  Bug: i => 0 is always true as `i` is an unsigned type.
for(i = num_of_bits-1; i >= 0; --i)  {
    printf("%d", arr[i]);
}

OP's repaired code.

float FloatAnalysis(float number) {
  assert(sizeof(float) == sizeof(unsigned));
  int arr[32] = {0};

  //float num_cpy = number;
  unsigned num_cpy;
  memcpy(&num_cpy, &number, sizeof num_cpy);  // copy the bit pattern to an unsigned

  // size_t num_of_bits = 0, i = 0;
  size_t num_of_bits = 32, i = 0;  // Always print 32 bits

  //while (0 != number) {
  //  num_cpy >>= 1;
  //  ++num_of_bits;
  //}
  //num_cpy = number;

  for (i = 0; i < num_of_bits; ++i) {
    arr[i] = (num_cpy & 1);
    num_cpy >>= 1;
  }

  // for(i = num_of_bits-1; i => 0; --i)
  for (i = num_of_bits; i-- > 0; ) { // Change test condition
    printf("%d", arr[i]);
  }
  printf("\n");

  // Some more output
  //      12345678901234567890123456789012
  printf("sEeeeeeeeMmmmmmmmmmmmmmmmmmmmmmm\n");
  printf("%a\n", number);
  return number;
}

int main() {
  FloatAnalysis(4.2f);
}

Output

01000000100001100110011001100110
sEeeeeeeeMmmmmmmmmmmmmmmmmmmmmmm
0x1.0cccccp+2

Print bits of float number does not work, can you explain me why?

2 Answers2

Use memcpy

Example

More reading