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.)