I have a class template that accepts a type T. It has a method. I want this method to return type T if it is const and T& if it is non-const.
template<typename T>
class C
{
static typename add_reference_if_non_const<T>::type method();
};
int main()
{
assert( is_same<result_of<C<int>::method()>::type, int&>::value );
assert( is_same<result_of<C<const int>::method()>::type, const int>::value );
}
How can I do it?
You want the return type of C<int const>::method() to be int const, but top-level cv qualifiers are ignored on function return types. In any case, since method() returns a copy of T, do you really care that you return T const and not T?
Given that, I think what you want is the following
using add_reference_if_non_const =
typename std::conditional<std::is_const<T>{},
typename std::remove_const<T>::type,
typename std::add_lvalue_reference<T>::type
>::type;
static add_reference_if_non_const method();
You can replace typename std::remove_const<T>::type with T if you want to return T const when T is a class type.
The next problem is with result_of which works with type arguments; what you have in the question is a function call of C::method. You need to use
result_of<decltype(&C<int>::method)()>::type
But since you need to use decltype anyway, you can completely do away with result_of.
decltype(C<int>::method())
Finally, you don't need to assert at runtime when you can make the checks at compile time using static_assert
static_assert( is_same<decltype(C<int>::method()), int&>::value, "");
static_assert( is_same<decltype(C<int const>::method()), int>::value, "" );
How about this:
template <typename T>
struct add_reference_if_non_const
{
using type = std::conditional<std::is_const<T>::value, T, std::add_lvalue_reference<T>>::type;
};
