cout uint8_t 作为整数而不是字符-C/C++问题

cout uint8_t as integers instead of chars(cout uint8_t 作为整数而不是字符)

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问题描述

#include<algorithm>
#include<iostream>
using namespace std;

int main()
{
    cout << (uint8_t)123 << endl;
}

这将输出 { ，因为 { 的 ASCII 是 123.

This will output { , since {'s ASCII is 123.

但我想得到 123 代替.我发现 cout <<(int)123< 会这样做，但我不愿意每次都将 uint_8 转换为 int .我可以配置 cout 来实现这一点吗?


But I want to get 123 instead. I found cout << (int)123 << endl; will do this, but I'm not willing to cast uint_8 to int every times. Can I configure cout to achieve this?
推荐答案
我绝对不会容忍我即将提出的解决方案.我也怀疑它可能不被标准所允许，但到目前为止我还不能证明这一点.如果有人可以向我提供表明不允许这样做的参考，那么我将删除此答案.无论如何，到目前为止，我的测试表明，在全局范围内简单地重载运算符似乎是可行的.
I definitely do not condone the solution I am about to suggest.  I also suspect that it may not be permitted by the standard, but I cannot prove it, as of yet.  If someone can provide me a reference that shows that it is not permitted, then I will delete this answer.  Anyway, my tests so far indicate that simply overloading the operator in the global scope seems to work.
#include <iostream>
#include <cstdint>

std::ostream & operator<<(std::ostream & os, std::uint8_t val)
{
    return os << static_cast<int>(val);
}

int main()
{
    std::uint8_t val = 123;
    std::cout << val;
}

我没想到这会起作用，但后来我意识到 operator<< 的 char/unsigned char/signed char 重载都是免费函数在 ADL 选取的 std 命名空间中.我想全局函数被认为比 ADL 函数更匹配，但我不确定这一点.
I wouldn't have thought this would work, but then I realized that the char/unsigned char/signed char overloads for operator<< are all free functions in the std namespace picked up by ADL.  And I guess global functions are considered a better match than ADL functions, but I'm not sure about that.

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