Diferencia entre char y char firmado en c ++?
Considere el siguiente código:
#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;
}
El resultado es :
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
Lo que significa queint
ysigned int
se consideran del mismo tipo, pero nochar
ysigned char
. Porqué es eso ?
Y si puedo transformar unchar
dentrosigned char
utilizandomake_signed
, como hacer lo contrario (transformar unsigned char
a unchar
)?