C / C ++将signed char转换为int [英] C/C++ packing signed char into int

问题描述

我需要将四个有符号字节打包成32位整数类型。
这是我想到的：

I have need to pack four signed bytes into 32-bit integral type. this is what I came up to:

int32_t byte(int8_t c) { return (unsigned char)c; }

int pack(char c0, char c1, ...) {
  return byte(c0) | byte(c1) << 8 | ...;
}

这是一个很好的解决方案吗？它是便携式的（不是在通信意义上）？

is this a good solution? Is it portable (not in communication sense)? is there a ready-made solution, perhaps boost?

问题我最关心的是将位从char转换为int时的位顺序。

issue I am mostly concerned about is bit order when converting of negative bits from char to int. I do not know what the correct behavior should be.

感谢

推荐答案

p>我喜欢Joey Adam的答案，除了它是用宏编写的（这在许多情况下导致真正的痛苦），如果'char'不是1字节宽，编译器不会给你警告。这是我的解决方案（基于Joey的）。

I liked Joey Adam's answer except for the fact that it is written with macros (which cause a real pain in many situations) and the compiler will not give you a warning if 'char' isn't 1 byte wide. This is my solution (based off Joey's).

inline uint32_t PACK(uint8_t c0, uint8_t c1, uint8_t c2, uint8_t c3) {
    return (c0 << 24) | (c1 << 16) | (c2 << 8) | c3;
}

inline uint32_t PACK(sint8_t c0, sint8_t c1, sint8_t c2, sint8_t c3) {
    return PACK((uint8_t)c0, (uint8_t)c1, (uint8_t)c2, (uint8_t)c3);
}

我省略了转换c0-> c3到uint32_t作为编译器应该处理这个为你移动时，我使用c风格的casts，因为他们将工作为c或c ++（OP标记为两者）。

I've omitted casting c0->c3 to a uint32_t as the compiler should handle this for you when shifting and I used c-style casts as they will work for either c or c++ (the OP tagged as both).

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