C语⾔uint8_t和char的区别,c–int8_t和uint8_t是char类型吗？鉴于这个C 11计划，我应该期待看到⼀个数字还是⼀个字母？还是没有期望？
#include
#include
int main()
{
int8_t i = 65;
std::cout << i;
}
标准是否指定此类型是否可以是字符类型？
解决⽅法:
根据C 0x FDIS(N3290)的§18.4.1[cstdint.syn],int8_t是⼀个可选的typedef，其指定如下：
namespace std {
typedef signed integer type int8_t; // optional
//...
} // namespace std
§3.9.1[basic.fundamental]陈述：
There are five standard signed integer types: “signed char”, “short int”, “int”, “long int”, and “long long int”. In
this list, each type provides at least as much storage as those preceding it in the list. There may also be implementation-defined extended signed integer types. The standard and extended signed integer types are collectively called signed integer types.
…
Types bool, char, char16_t, char32_t, wchar_t, and the signed and unsigned integer types are collectively called integral types. A synonym for integral type is integer type.
§3.9.1还规定：
In any particular implementation, a plain char object can take on either the same values as a signed char or an unsigned char; which one is implementation-defined.
很有可能得出结论,int8_t可能是char的typedef，前提是char对象采⽤有符号值；但是，情况并⾮如此，因为char不在有符号整数类型列表中(标准和可能扩展的有符号整数类型).另请参见std :: make_unsigned和std :: make_signed上的Stephan T. Lavavej’s comments.
因此,int8_t是signed char的typedef，或者是扩展的有符号整数类型，其对象恰好占⽤8位存储.
但是，要回答你的问题，你不应该做出假设.因为已经定义了x.operator<
>模板< class traits> basic_ostream<炭，性状>&安培；如果int8_t是签名字符的完全匹配(即signed char的typedef)，则将调⽤operator&,signed char)模板.
>否则,int8_t将被提升为int和basic_ostream< charT,traits>&将调⽤operator
在std :: cout<
>模板< class traits> basic_ostream<炭，性状>&安培；如果uint8_t是unsigned char的完全匹配，则将调⽤operator&,unsigned char)模板.
namespace是干嘛的>否则，因为int可以表⽰所有uint8_t值,uint8_t将被提升为int和basic_ostream< charT,traits>&将调⽤operator 如果您总是想要打印⼀个⾓⾊，最安全，最明确的选择是：
std::cout << static_cast(i);
如果你总想打印⼀个数字：
std::cout << static_cast(i);
标签：c,c11,language-lawyer,iostream,standard-library

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