Doing a static_assert that a template type is another template

Question

How do I static_assert like this? Maybe Boost supports it if not C++ or new features in C++11?

template<T>
struct foo {};

template<FooType>
struct bar {
  static_assert(FooType is indeed foo<T> for some T,"failure"); //how?
};

Answer 1

You could do something along these lines. Given a trait that can verify whether a class is an instantiation of a class template:

#include <type_traits>

template<typename T, template<typename> class TT>
struct is_instantiation_of : std::false_type { };

template<typename T, template<typename> class TT>
struct is_instantiation_of<TT<T>, TT> : std::true_type { };

Use it as follows in your program:

template<typename T>
struct foo {};

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

int main()
{
    bar<int> b; // ERROR!
    bar<foo<int>> b; // OK!
}

If you want, you could generalize this to detect whether a class is an instance of a template with any number of (type) parameters, like so:

#include <type_traits>

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

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

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

You would then use it this way in your program:

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

int main()
{
    bar<int> b; // ERROR!
    bar<foo<int>> b; // OK!
}

Here is a live example.

Answer 2

Some small improvements over the other answers:

• the name actually makes sense regarding the order of the parameters
• handles const, volatile, and reference types properly via std::decay
• implements C++14-style _v constexpr variable
• accepts an arbitrary number of types to test against (tests for ALL)

I have intentionally not put the std::decay_t on the is_template_for_v because a type trait should work identically regardless of whether it is called with the _v suffix or not.

This does require C++17 for std::conjunction, but you can either remove the variadic feature or implement your own conjunction using c++11/14.

template<template<class...> class tmpl, typename T>
struct _is_template_for : public std::false_type {};

template<template<class...> class tmpl, class... Args>
struct _is_template_for<tmpl, tmpl<Args...>> : public std::true_type {};

template<template<class...> class tmpl, typename... Ts>
using is_template_for = std::conjunction<_is_template_for<tmpl, std::decay_t<Ts>>...>;

template<template<class...> class tmpl, typename... Ts>
constexpr bool is_template_for_v = is_template_for<tmpl, Ts...>::value;

Usage:

static_assert(is_template_for_v<std::vector, std::vector<int>>); // doesn't fire

Answer 3

As someone else wrote,

template<typename T, template<typename...> class TT>
struct is_specialization_of : std::false_type { };

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

However, beware that this works only for template classes whose template parameters are all typenames! Presented with

typedef std::array<int, 42> MyArray;
static_assert(is_specialization_of<MyArray, std::array>::value, "");

it will simply fail to compile at all.

I believe C++11/C++14/C++17 currently have no way to deal with this limitation.