Question8.10：I'm starting to think about multinational character sets。。。

那个谁

Q: I'm starting to think about multinational character sets, and I'm worried about the implications of making sizeof(char) be 2 so that 16-bit character sets can be represented.
A: If type char were made 16 bits, sizeof(char) would still be 1, and CHAR_BIT in <limits.h> would be 16, and it would simply be impossible to declare (or allocate with malloc) a single 8-bit object.

Traditionally, a byte is not necessarily 8 bits, but merely a smallish region of memory, usually suitable for storing one character. The C Standard follows this usage, so the bytes used by malloc and sizeof can be more than 8 bits.

To allow manipulation of multinational character sets without requiring an expansion of type char, ANSI/ISO C defines the ``wide&#39;&#39; character type wchar_t, and corresponding wide string literals, and functions for manipulating and converting strings of wide characters.

See also question 7.8.

那个谁

问题：我开始考虑多国的字符集的问题，我想将sizeof（char）定义为2，那么16位字符集就可以被表示了。
回答：如果char型被认为是16位，sizeof（char）将仍然为1，并且CHAR_BIT在<limits.h>中为16。那么将不可能声明（或者是用malloc分配）一个简单的8位对象了。
传统上，一个字节并不一定必须是8位，它仅仅是内存中的一小段区域，经常用来存储一个字符的大小。标准C也使用了这种用法，因此malloc和sizeof都可以使用的字节都是8位以上。
为了允许在没有扩展字符类型的情况下操作多国的字符集，ANSI/ISO C定义了“宽”字符型wchar_t，和相应的宽字符串，以及用来操作和转换宽字符串的函数。
参考问题7.8。