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

Question

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; } 

Answer 1

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.)