以整数而不是char的形式输出uint8_t [英] cout uint8_t as integers instead of chars

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.

但是我想改为获取 123 .我发现 cout<<(int)123<<endl; 将执行此操作，但是我不愿意每次将 uint_8 强制转换为 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

我绝对不容忍我要提出的解决方案.我还怀疑该标准可能不允许使用它，但是到目前为止我还不能证明它.如果有人可以向我提供表明不允许的参考，那么我将删除此答案.无论如何，到目前为止，我的测试表明，简单地在全局范围内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;
}

我本来以为这行不通，但是后来我意识到 operator<< 的 char/unsigned char/signed char 重载都是自由函数在ADL选取的 std 名称空间中.而且我认为全局函数比ADL函数更匹配，但是我不确定.

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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