c++floating-pointendiannessportabilitysingle-precision

Building a 32-bit float out of its 4 composite bytes


I'm trying to build a 32-bit float out of its 4 composite bytes. Is there a better (or more portable) way to do this than with the following method?

#include <iostream>

typedef unsigned char uchar;

float bytesToFloat(uchar b0, uchar b1, uchar b2, uchar b3)
{
    float output;

    *((uchar*)(&output) + 3) = b0;
    *((uchar*)(&output) + 2) = b1;
    *((uchar*)(&output) + 1) = b2;
    *((uchar*)(&output) + 0) = b3;

    return output;
}

int main()
{
    std::cout << bytesToFloat(0x3e, 0xaa, 0xaa, 0xab) << std::endl; // 1.0 / 3.0
    std::cout << bytesToFloat(0x7f, 0x7f, 0xff, 0xff) << std::endl; // 3.4028234 × 10^38  (max single precision)

    return 0;
}

Solution

  • You could use a memcpy (Result)

    float f;
    uchar b[] = {b3, b2, b1, b0};
    memcpy(&f, &b, sizeof(f));
    return f;
    

    or a union* (Result)

    union {
      float f;
      uchar b[4];
    } u;
    u.b[3] = b0;
    u.b[2] = b1;
    u.b[1] = b2;
    u.b[0] = b3;
    return u.f;
    

    But this is no more portable than your code, since there is no guarantee that the platform is little-endian or the float is using IEEE binary32 or even sizeof(float) == 4.

    (Note*: As explained by @James, it is technically not allowed in the standard (C++ §[class.union]/1) to access the union member u.f.)