为什么std::( i)ostream将带符号/无符号字符视为文本而不是整数?

问题描述

此代码没有执行应做的事情:

This code doesn't do what it's supposed to do:

#include <iostream>
#include <cstdint>

int main()
{
    uint8_t small_integer;
    std::cin >> small_integer;
    std::cout << small_integer;
}

原因很简单:uint8_t是unsigned char的typedef，流将此类型视为文本:
Visual C ++ 2015实现

The reason is simple: uint8_t is a typedef for unsigned char and streams treat this type as a text:
Visual C++ 2015 implementation

template<class _Traits> inline
    basic_istream<char, _Traits>& operator>>(
        basic_istream<char, _Traits>& _Istr, unsigned char& _Ch)
    {    // extract an unsigned char
    return (_Istr >> (char&)_Ch);
    }

和类似的代码，其中operator <<强制转换为char.

And a similar code with cast to char for operator <<.

我的问题:

1. 标准是否要求这种行为(流操作符将有符号/无符号字符视为字符类型，而不是整数)?如果是，则:

1. Is this behavior (streaming operators treating signed / unsigned char as character type and not an integer) required by the standard?If it is then:

1. 这种违反直觉的语义背后的原理是什么?
2. 应该认为这是一个缺陷吗，是否有提议来更改这种语义?

我可能应该添加一些解释，为什么我认为这违反直觉.尽管类型名称包含单词char，但是signed或unsigned部分指定了特定的整数语义，并且这些类型通常用作字节大小的整数.甚至标准也通过它们定义了int8_t/uint8_t.

I should probably add a little explanation why I consider it counterintuitive.Although the type name contains the word char, the signed or unsigned part specify particular integer semantic and those types are generally used as byte sized integers. Even the standard defines int8_t / uint8_t through them.

UPD::问题是有关unsigned char和signed char的流运算符重载的行为.

UPD: The question is about behavior of streaming operator overloads for unsigned char and signed char.

推荐答案

标准(n3797)表示以下内容:

The standard (n3797) says the following:

template<class charT, class traits>
basic_istream<charT,traits>& operator>>(basic_istream<charT,traits>& in, charT& c);

template<class traits>
basic_istream<char,traits>& operator>>(basic_istream<char,traits>& in, unsigned char& c);

template<class traits>
basic_istream<char,traits>& operator>>(basic_istream<char,traits>& in, signed char& c);

27.7.3.6.4字符插入器功能模板

// specialization
template<class traits>
basic_ostream<char,traits>& operator<<(basic_ostream<char,traits>& out, char c);

// signed and unsigned
template<class traits>
basic_ostream<char,traits>& operator<<(basic_ostream<char,traits>& out, signed char c);

template<class traits>
basic_ostream<char,traits>& operator<<(basic_ostream<char,traits>& out, unsigned char c);

第一个问题的答案是:是的，标准要求operator >>和operator <<对于char，unsigned char和signed char的行为完全相同，即它们读/写单个字符，而不是整数.不幸的是，标准并不能解释原因.我希望有人能阐明2和3.

So the answer to the first question: yes, the standard requires that operator >> and operator << behave exactly the same for char, unsigned char and signed char, that is they read / write a single character, not an integer. Unfortunately, standard doesn't explain why. I hope someone will shed light on 2 and 3.

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