I want to determine the underlying template of a template parameter by using a combination of a template alias and template specializations. The follwing code compiles fine on gcc 4.8, 6.2.1 but not on clang 3.5, 3.8.
#include <iostream>
template <typename T> struct First {};
template <typename T> struct Second {};
template <template <typename> class F, typename T> struct Foo {};
template <typename T> struct Foo<First, T>
{
void f() { std::cout << __PRETTY_FUNCTION__ << std::endl; }
};
template <typename T> struct Foo<Second, T>
{
void f() { std::cout << __PRETTY_FUNCTION__ << std::endl; }
};
template <typename F, typename T> struct Resolution {};
template <typename T> struct Resolution<First<T>, T>
{
template <typename P> using type = First<P>;
};
template <typename T> struct Resolution<Second<T>, T>
{
template <typename P> using type = Second<P>;
};
int main()
{
Foo<Resolution<First<int>, int>::type, float> my_foo;
my_foo.f(); // main.cpp:34:12: error: no member named 'f' in 'Foo<Resolution<First<int>, int>::type, float>'
return 0;
}
Which behavior is standard conformant?
Resolution<First<int>, int>::type
(which despite the name is a template) againstFirst
in theFoo<First, T>
partial specialization. – Mountaineer