用 4 个复合字节构建一个 32 位浮点数

沃梦达教程 C/C++问题

2022-01-01

Building a 32-bit float out of its 4 composite bytes(用 4 个复合字节构建一个 32 位浮点数)

本文介绍了用 4 个复合字节构建一个 32 位浮点数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在尝试用它的 4 个复合字节构建一个 32 位浮点数.有没有比使用以下方法更好(或更便携)的方法?

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.


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. 
(注意*:如 @James，技术上不允许在标准 (C++ §[class.union]/1) 中访问联合成员 uf.)
(Note*: As explained by @James, it is technically not allowed in the standard (C++ §[class.union]/1) to access the union member u.f.)

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