C++ cout hex format

Question

i am a c coder, new to c++.

i try to print the following with cout with strange output. Any comment on this behaviour is appreciated.

#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);
}

output:

 Value of x  ÿ  hexadecimal
 Value of x ff by printf

Answer 1

<< handles char as a 'character' that you want to output, and just outputs that byte exactly. The hex only applies to integer-like types, so the following will do what you expect:

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

Billy ONeal's suggestion of static_cast would look like this:

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

Answer 2

You are doing the hex part correctly, but x is a character, and C++ is trying to print it as a character. You have to cast it to an integer.

#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);
}