用 4 个复合字节构建一个 32 位浮点数

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;
}

推荐答案

你可以使用 memcpy (结果)

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

或联合*(结果)

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;

但这并不比你的代码更便携，因为不能保证平台是 little-endian 或 float 使用 IEEE binary32 甚至 sizeof(float) ==4.