C++ 中是否有实际的 8 位整数数据类型
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/35246212/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Is there an actual 8-bit integer data type in C++
提问by tej-kweku
In c++, specifically the cstdint header file, there are types for 8-bit integers which turn out to be of the char data type with a typedef. Could anyone suggest an actual 8-bit integer type?
在 c++ 中，特别是 cstdint 头文件中，有一些 8 位整数类型，它们是带有 typedef 的 char 数据类型。谁能建议一个实际的 8 位整数类型？
回答by dlmeetei
Yes, you are right. int8_tand uint8_tare typedefto charon platforms where 1 byte is 8 bits. On platforms where it is not, appropriate definition will be given.
你是对的。int8_t并且uint8_t是typedef到char平台上，其中1个字节为8位。在没有它的平台上，将给出适当的定义。
Following answer is based on assumption that char is 8 bits
以下答案基于假设 char 是 8 bits
charholds 1 byte, which may be signedor unsignedbased on implementation.
char持有 1 个字节，这可能是signed或unsigned基于实现。
So int8_tis signed charand uint8_tis unsigned char, but this will be safe to use int8_t/uint8_t as actual 8-bit integer without relying too much on the implementation.
所以，int8_t是signed char和uint8_t是unsigned char，但是这将是安全的使用中int8_t / uint8_t作为实际的8位整数，而不过分依赖于实现。
For a implementer's point of view, typedeffing where char is 8 bits makes sense.
对于实现者的观点，typedeffing 其中 char 是 8 位是有意义的。
Having seen all this, It is safe to use int8_tor uint8_tas real 8 bit integer.
看到这一切后，可以安全地使用int8_toruint8_t作为真正的 8 位整数。
