c++ - Using a template alias instead of a template within a template

Question

From a previous question:

Doing a static_assert that a template type is another template

Andy Prowl provided me with this code that allows me to static_assert that a template type is another template type:

template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of : public std::false_type { };

template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of<TT, TT<Ts...>> : public std::true_type { };

template<typename T>
struct foo {};

template<typename FooType>
struct bar {
  static_assert(is_instantiation_of<foo,FooType>::value, ""); //success
};

int main(int,char**)
{
  bar<foo<int>> b; //success
  return 0;
}

This works great.

But if I change the code like this to use an alias of foo, things go bad:

template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of : public std::false_type { };

template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of<TT, TT<Ts...>> : public std::true_type { };

template<typename T>
struct foo {};

//Added: alias for foo
template<typename T>
using foo_alt = foo<T>;

template<typename FooType>
struct bar {
  //Changed: want to use foo_alt instead of foo here
  static_assert(is_instantiation_of<foo_alt,FooType>::value, ""); //fail
};

int main(int,char**) {
  //both of these fail:
  bar<foo<int>> b;
  bar<foo_alt<int>> b2;

  return 0;
}

Can this be solved?

Answer 1

No, it cannot be solved (not without changing the design significantly at least). The problem is that template alias names are not deduced, as mentioned in paragraph 14.5.7/2 of the C++11 Standard:

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 ]

The paragraph also provides an example:

[ 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<template<class> class TT>
void f(TT<int>);
f(v); // error: Vec not deduced                          <=== Relevant

...

—end example ]

In your concrete case, the problem is that when trying to match the partial specialization, the compiler won't deduce that your type is an instantiation of foo_alt (since foo_alt is the name of an alias template), and the primary template gets picked.

If you want to use alias templates, you will have to give up a bit of genericity and create a type trait specific for foo:

#include <type_traits>

template<typename T>
struct foo {};

template<typename T>
struct is_instantiation_of_foo : std::false_type { };

template<typename...Ts>
struct is_instantiation_of_foo<foo<Ts...>> : std::true_type { };

Which you could then use this way:

template<typename FooType>
struct bar {
  static_assert(is_instantiation_of_foo<FooType>::value, ""); //fail
};

Now, none of the assertions in the following program will fire:

template<typename T>
using foo_alt = foo<T>;

int main(int,char**) {
  // None of these fail:
  bar<foo<int>> b;
  bar<foo_alt<int>> b2;

  return 0;
}

Here is a live example.