I have been reading about removing reference of a type, here.
It gives the following example:
#include <iostream> // std::cout
#include <type_traits> // std::is_same
template<class T1, class T2>
void print_is_same() {
std::cout << std::is_same<T1, T2>() << '\n';
}
int main() {
std::cout << std::boolalpha;
print_is_same<int, int>();
print_is_same<int, int &>();
print_is_same<int, int &&>();
print_is_same<int, std::remove_reference<int>::type>(); // Why not typename std::remove_reference<int>::type ?
print_is_same<int, std::remove_reference<int &>::type>();// Why not typename std::remove_reference<int &>::type ?
print_is_same<int, std::remove_reference<int &&>::type>();// Why not typename std::remove_reference<int &&>::type ?
}
The type
s in the std::remove_reference
traits are dependent types.
Possible implementation
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;};
But why does it not use typename std::remove_reference</*TYPE*/>::type
?
The
type
s in thestd::remove_reference
traits are dependent types.
No, they are not dependent names here. The template arguments have been specified explicitly as int
, int&
and int&&
. Therefore, the types are known at this point.
On the other hand, if you use std::remove_reference
with a template parameter, e.g.
template <typename T>
void foo() {
print_is_same<int, typename std::remove_reference<T>::type>();
}
then you have to use typename
to tell that std::remove_reference<T>::type
is a type as your expression now depends on the template parameter T
.