stdint 的 int8_t 可以存在于没有 8 位字节的体系结构上吗? [英] Can stdint's int8_t exist on an architecture that does not have 8-bit bytes?
问题描述
似乎这样的架构会排除 int8_t(在 stdint.h 中定义)类型的存在，因为根据我的理解，C 不能创建小于一个 CHAR_BIT.
It would seem that such architectures would preclude the existence of an int8_t (defined in stdint.h) type since C, from my understanding, cannot create datatypes smaller than a CHAR_BIT.
也就是说，IEEE stdint.h def似乎要求存在这种类型(以及其他类型)，只允许在不支持它的架构上不存在 64 位.
That said, the IEEE stdint.h def seems to require that such a type exist (along with others), only allowing for the 64-bit to not exist on architectures that do not support it.
我错过了什么吗?
推荐答案
正如@JasonD 在下面的评论中指出的那样，链接页面在结尾处声明；
As @JasonD points out in the comments below, the linked page states at the end;
添加 int8_t 的结果是:
As a consequence of adding int8_t, the following are true:
一个字节正好是 8 位.
A byte is exactly 8 bits.
{CHAR_BIT} 的值为 8，{SCHAR_MAX} 的值为 127，{SCHAR_MIN} 的值为 -128，{UCHAR_MAX} 的值为 255.
{CHAR_BIT} has the value 8, {SCHAR_MAX} has the value 127, {SCHAR_MIN} has the value -128, and {UCHAR_MAX} has the value 255.
换句话说，链接的 IEEE 页面不适用于字节长度不为 8 的架构.这符合需要 8 位字符的 POSIX.
In other words, the linked IEEE page does not apply to architectures with other byte lengths than 8. This is in line with POSIX which requires 8 bit char.
-- 编辑前--
<打击>解释在您链接到的页面的注释中；
整数类型的宽度"是用于在纯二进制系统中存储其值的位数；实际类型可能使用更多位(例如，28 位类型可以存储在 32 位实际存储中)
The "width" of an integer type is the number of bits used to store its value in a pure binary system; the actual type may use more bits than that (for example, a 28-bit type could be stored in 32 bits of actual storage)
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