如何通过cout将字符输出为整数？ [英] How to output a character as an integer through cout?

问题描述

#include <iostream>

using namespace std;

int main()
{  
    char          c1 = 0xab;
    signed char   c2 = 0xcd;
    unsigned char c3 = 0xef;

    cout << hex;
    cout << c1 << endl;
    cout << c2 << endl;
    cout << c3 << endl;
}

我希望输出如下：

I expected the output are as follows:

ab
cd
ef

b $ b

但是，我什么都没有。

Yet, I got nothing.

我猜这是因为cout总是把'char'，'signed char'和'unsigned char'字符而不是8位整数。但是，'char'，'signed char'和'unsigned char'都是整数类型。

I guess this is because cout always treats 'char', 'signed char', and 'unsigned char' as characters rather than 8-bit integers. However, 'char', 'signed char', and 'unsigned char' are all integral types.

所以我的问题是： cout？

So my question is: How to output a character as an integer through cout?

PS：static_cast（...）很丑陋，需要更多的工作来修剪额外的位。

PS: static_cast(...) is ugly and needs more work to trim extra bits.

推荐答案

我知道这个问题是很老的，但无论如何...

I know this question is pretty old but anyway...

char a = 0xab;
cout << +a; // promotes x to a type printable as a number, regardless of type

类型为一元+运算符提供普通语义。如果您要定义一个表示数字的类，则要提供一个带有规范语义的一元+运算符，创建一个运算符+（），它通过value或者reference-to-const返回* this。

This works as long as the type provides a unary + operator with ordinary semantics. If you are defining a class that represents a number, to provide a unary + operator with canonical semantics, create an operator+() that simply returns *this either by value or by reference-to-const.

来源： Parashift.com - 如何打印一个字符作为数字？如何打印一个char *，以便输出显示指针的数字值？

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