17 Templates [temp]

17.5 Template declarations [temp.decls]

17.5.7 Alias templates [temp.alias]

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]