炭尺寸混乱 [英] char size confusion

问题描述

我知道1个char = 1个字节= 8位（32位系统）。

As per I know that 1 char = 1 byte = 8 bits(32 bit system).

char c=0xffff0000;  //wrong

那么为什么 char 8位以及 8位长度的文件中的每个字符。

then why char allow just 8 bits and also every character in a file also of 8 bit length.

推荐答案

否。 size的char是从定义1。但这并不意味着它总是占用32位/ 8位。

No. The sizeof char is by definition 1. But this does not mean that it occupies 32-bits/8-bits always.

$ 3.9.1 / 1-声明为字符（char）的对象应该是大的
，足以存储
实现的基本字符集的任何成员。

$3.9.1/1- "Objects declared as characters (char) shall be large enough to store any member of the implementation’s basic character set."

似乎有一个混乱，一个字节是8位。

There appears to be a confusion that a byte is 8-bits. The C++ Standard does not mandate this however.

以下是如何在标准$ 1.7 / 1中定义字节

Here's how byte is defined in the Standard $1.7/1

C
+ +内存模型中的基本存储单元是字节。一个字节至少足够大以包含
基本执行
字符集的任何成员，并由一个
连续的位序列
编号组成是
实现定义的。

很明显，一个字节不需要总是8位。

As is clear, a byte need not be always 8-bits.

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