系统，1个字节！= 8位？ [英] System where 1 byte != 8 bit?

问题描述

所有时间我喜欢读

不靠1个字节是大小为8位

don't rely on 1 byte being 8 bit in size

使用 CHAR_BIT 而不是8作为常数比特和字节之间的转换

use CHAR_BIT instead of 8 as a constant to convert between bits and bytes

等等。什么是真正的生活系统今天在那里，这也是如此？
<子>（我不知道是否有++对此C和C之间的差异，或者如果它实际上是语言无关的，请重新标记，如果neccessary。）

et cetera. What real life systems are there today, where this holds true? _{(I'm not sure if there are differences between C and C++ regarding this, or if it's actually language agnostic. Please retag if neccessary.)}

推荐答案

在老机器，codeS小于8位是相当普遍的，但其中大部分都已经死了好几年了。

On older machines, codes smaller than 8 bits were fairly common, but most of those have been dead and gone for years now.

C和C ++已经授权一个的最低8位的为字符，至少早在C89标准。

C and C++ have mandated a minimum of 8 bits for char, at least as far back as the C89 standard. . They treat "char" and "byte" as essentially synonymous

有，但是，当前的机器（主要的DSP），其中最小的类型是大于8位 - 最低12，14，或甚至16比特是相当普遍的。 Windows CE的的确大致相同：它的最小类型（至少与微软的编译器）是16位。他们这样做的不的，但是，治疗字符 16位 - 相反，他们采取的只是不支持类型（不合格）的方法名为字符的。

There are, however, current machines (mostly DSPs) where the smallest type is larger than 8 bits -- a minimum of 12, 14, or even 16 bits is fairly common. Windows CE does roughly the same: its smallest type (at least with Microsoft's compiler) is 16 bits. They do not, however, treat a char as 16 bits -- instead they take the (non-conforming) approach of simply not supporting a type named char at all.

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