template< class T >
struct remove_reference;
If the type T is a reference type, provides the member typedef type which is the type referred to by T. Otherwise type is T.
If the program adds specializations for std::remove_reference, the behavior is undefined.
type the type referred by T or T if it is not a reference [edit] Helper types
template< class T >
using remove_reference_t = typename remove_reference<T>::type;
template<class T> struct remove_reference { typedef T type; };
template<class T> struct remove_reference<T&> { typedef T type; };
template<class T> struct remove_reference<T&&> { typedef T type; };[edit] Example
#include <iostream>
#include <type_traits>
int main()
{
std::cout << std::boolalpha;
std::cout << "std::remove_reference<int>::type is int? "
<< std::is_same<int, std::remove_reference<int>::type>::value << '\n';
std::cout << "std::remove_reference<int&>::type is int? "
<< std::is_same<int, std::remove_reference<int&>::type>::value << '\n';
std::cout << "std::remove_reference<int&&>::type is int? "
<< std::is_same<int, std::remove_reference<int&&>::type>::value << '\n';
std::cout << "std::remove_reference<const int&>::type is const int? "
<< std::is_same<const int,
std::remove_reference<const int&>::type>::value << '\n';
}
Output:
std::remove_reference<int>::type is int? true std::remove_reference<int&>::type is int? true std::remove_reference<int&&>::type is int? true std::remove_reference<const int&>::type is const int? true[edit] See also
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