C++ cout十六进制格式

Question

我是ac编码器,是c ++的新手.

我尝试用奇怪的输出cout打印以下内容.对此行为的任何评论表示赞赏.

#include<iostream>
using namespace std;

int main()
{
        unsigned char x = 0xff;

        cout << "Value of x  " << hex<<x<<"  hexadecimal"<<endl;

        printf(" Value of x %x by printf", x);
}

输出:

 Value of x  ÿ  hexadecimal
 Value of x ff by printf

Answer 1

<<处理char您想要输出的"字符",并且只输出该字节.该hex只适用于类似于整数类型,所以下面会做你所期望的:

cout << "Value of x  " << hex << int(x) << "  hexadecimal" << endl;

Billy ONeal的建议static_cast看起来像这样:

cout << "Value of x  " << hex << static_cast<int>(x) << "  hexadecimal" << endl;

Answer 2

您正确地执行了十六进制部分，但 x 是一个字符，而 C++ 试图将其打印为一个字符。您必须将其强制转换为整数。

#include<iostream>
using namespace std;

int main()
{
        unsigned char x = 0xff;

        cout << "Value of x  " << hex<<static_cast<int>(x)<<"  hexadecimal"<<endl;

        printf(" Value of x %x by printf", x);
}