A template-declaration in which the declaration is an alias-declaration declares the identifier to be an alias template. An alias template is a name for a family of types. The name of the alias template is a template-name.
When a template-id refers to the specialization of an alias template, it is equivalent to the associated type obtained by substitution of its template-arguments for the template-parameters in the type-id of the alias template. [ Note: An alias template name is never deduced. — end note ] [ Example:
template<class T> struct Alloc { /* ... */ };
template<class T> using Vec = vector<T, Alloc<T>>;
Vec<int> v; // same as vector<int, Alloc<int>> v;
template<class T>
void process(Vec<T>& v)
{ /* ... */ }
template<class T>
void process(vector<T, Alloc<T>>& w)
{ /* ... */ } // error: redefinition
template<template<class> class TT>
void f(TT<int>);
f(v); // error: Vec not deduced
template<template<class,class> class TT>
void g(TT<int, Alloc<int>>);
g(v); // OK: TT = vector
— end example ]
However, if the template-id is dependent, subsequent template argument substitution still applies to the template-id. [ Example:
template<typename...> using void_t = void;
template<typename T> void_t<typename T::foo> f();
f<int>(); // error, int does not have a nested type foo
— end example ]
The type-id in an alias template declaration shall not refer to the alias template being declared. The type produced by an alias template specialization shall not directly or indirectly make use of that specialization. [ Example:
template <class T> struct A;
template <class T> using B = typename A<T>::U;
template <class T> struct A {
typedef B<T> U;
};
B<short> b; // error: instantiation of B<short> uses own type via A<short>::U
— end example ]