如何通过cout将字符输出为整数？

Question

#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;
}

我预计输出如下:

ab
cd
ef

然而,我一无所获.

我想这是因为cout总是将'char','signed char'和'unsigned char'视为字符而不是8位整数.但是,'char','signed char'和'unsigned char'都是完整的类型.

所以我的问题是:如何通过cout将字符输出为整数？

PS:static_cast(...)很难看,需要更多的工作来修剪额外的比特.

Answer 1

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

只要该类型为+普通语义提供一元运算符,这就可以工作.如果要定义一个表示数字的类,要为一元+运算符提供规范语义,请创建一个operator+()只返回*this值或引用到const.

来源:Parashift.com - 如何将字符作为数字打印？如何打印char*以便输出显示指针的数值？

Answer 2

将它们转换为整数类型,(和bitmask适当!)即:

#include <iostream>

using namespace std;

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

    cout << hex;
    cout << (static_cast<int>(c1) & 0xFF) << endl;
    cout << (static_cast<int>(c2) & 0xFF) << endl;
    cout << (static_cast<unsigned int>(c3) & 0xFF) << endl;
}

Answer 3

也许这个:

char c = 0xab;
std::cout << (int)c;

希望能帮助到你.