什么是int8_t如果机器有> 8位/字节？ [英] What is int8_t if a machine has > 8 bits per byte?

问题描述

我在阅读 C ++常见问题，并说

C ++语言保证一个字节必须至少有8位

那么，< cstdint> 类型是什么意思？

我应该使用 int8_t 或 char 为什么？

intX_t （即确切的宽度整数类型）typedefs作为可选。

如果你想要一个字节数组，你应该使用 char 如 sizeof char （和 signed char 和 unsigned char ）定义为总是1个字节。

I was reading the C++ FAQ and it says

The C++ language guarantees a byte must always have at least 8 bits

So what does that mean for the <cstdint> types?

Side question - if I want an array of bytes should I use int8_t or char and why?

解决方案

C++ (and C as well) defines intX_t (i.e. the exact width integer types) typedefs as optional. So, it just won't be there if there is no addressable unit that's exactly 8-bit wide.

If you want an array of bytes, you should use char, as sizeof char (and signed char and unsigned char) is well-defined to always be 1 byte.

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