Question

Code:

#include <stdio.h>

int main()
{
    printf(
            " %f, %u, %x,\n", 1.0f, 1.0f, 1.0f);
    return 0;
}

Output: 1.000000, 1072693248, 0,

Code:

#include <stdio.h>

int main()
{
    printf(
            " %x, %f, %u,\n", 1.0f, 1.0f, 1.0f);
    return 0;
}

Output: 3ff00000, 0.000000, 0

Code:

#include <stdio.h>

int main()
{
    printf(
            " %x, %u, %f,\n", 1.0f, 1.0f, 1.0f);
    return 0;
}

Output: 3ff00000, 0, 1.000000

Is this just an issue with the number of bytes that %u and %x consume, and how do I get the values to become consistent?

Answer 1

Passing arguments whose type does not match the corresponding component of the format string will lead to undefined results, as you’ve noticed. In particular, "%x" and "%u" both expect values of type (unsigned) int. Passing a float (which will often actually be represented as a double or long double depending on the ABI) is semantically incorrect.

Your compiler should be warning you about this – if not, make sure you’ve got all warnings enabled (-Wall for GCC).

If you want 1.0f as an integer, just cast:

printf(" %x, %u, %f\n", (unsigned)1.0f, (unsigned)1.0f, 1.0f);

If you’re trying to obtain the binary representations, try something like this:

float a = 1.0f;
printf(" %x, %u, %f\n", *((unsigned*)&a), *((unsigned*)&a), a);