c++11 - Difference between char and signed char in c++?

Question

Consider the following code :

#include <iostream>
#include <type_traits>

int main(int argc, char* argv[])
{
    std::cout<<"std::is_same<int, int>::value = "<<std::is_same<int, int>::value<<std::endl;
    std::cout<<"std::is_same<int, signed int>::value = "<<std::is_same<int, signed int>::value<<std::endl;
    std::cout<<"std::is_same<int, unsigned int>::value = "<<std::is_same<int, unsigned int>::value<<std::endl;
    std::cout<<"std::is_same<signed int, int>::value = "<<std::is_same<signed int, int>::value<<std::endl;
    std::cout<<"std::is_same<signed int, signed int>::value = "<<std::is_same<signed int, signed int>::value<<std::endl;
    std::cout<<"std::is_same<signed int, unsigned int>::value = "<<std::is_same<signed int, unsigned int>::value<<std::endl;
    std::cout<<"std::is_same<unsigned int, int>::value = "<<std::is_same<unsigned int, int>::value<<std::endl;
    std::cout<<"std::is_same<unsigned int, signed int>::value = "<<std::is_same<unsigned int, signed int>::value<<std::endl;
    std::cout<<"std::is_same<unsigned int, unsigned int>::value = "<<std::is_same<unsigned int, unsigned int>::value<<std::endl;
    std::cout<<"----"<<std::endl;
    std::cout<<"std::is_same<char, char>::value = "<<std::is_same<char, char>::value<<std::endl;
    std::cout<<"std::is_same<char, signed char>::value = "<<std::is_same<char, signed char>::value<<std::endl;
    std::cout<<"std::is_same<char, unsigned char>::value = "<<std::is_same<char, unsigned char>::value<<std::endl;
    std::cout<<"std::is_same<signed char, char>::value = "<<std::is_same<signed char, char>::value<<std::endl;
    std::cout<<"std::is_same<signed char, signed char>::value = "<<std::is_same<signed char, signed char>::value<<std::endl;
    std::cout<<"std::is_same<signed char, unsigned char>::value = "<<std::is_same<signed char, unsigned char>::value<<std::endl;
    std::cout<<"std::is_same<unsigned char, char>::value = "<<std::is_same<unsigned char, char>::value<<std::endl;
    std::cout<<"std::is_same<unsigned char, signed char>::value = "<<std::is_same<unsigned char, signed char>::value<<std::endl;
    std::cout<<"std::is_same<unsigned char, unsigned char>::value = "<<std::is_same<unsigned char, unsigned char>::value<<std::endl;
    return 0;
}

The result is :

std::is_same<int, int>::value = 1
std::is_same<int, signed int>::value = 1
std::is_same<int, unsigned int>::value = 0
std::is_same<signed int, int>::value = 1
std::is_same<signed int, signed int>::value = 1
std::is_same<signed int, unsigned int>::value = 0
std::is_same<unsigned int, int>::value = 0
std::is_same<unsigned int, signed int>::value = 0
std::is_same<unsigned int, unsigned int>::value = 1
----
std::is_same<char, char>::value = 1
std::is_same<char, signed char>::value = 0
std::is_same<char, unsigned char>::value = 0
std::is_same<signed char, char>::value = 0
std::is_same<signed char, signed char>::value = 1
std::is_same<signed char, unsigned char>::value = 0
std::is_same<unsigned char, char>::value = 0
std::is_same<unsigned char, signed char>::value = 0
std::is_same<unsigned char, unsigned char>::value = 1 

Which means that int and signed int are considered as the same type, but not char and signed char. Why is that ?

And if I can transform a char into signed char using make_signed, how to do the opposite (transform a signed char to a char) ?

Answer 1

It's by design, C++ standard says char, signed char and unsigned char are different types. I think you can use static cast for transformation.