How to detect if a type is one of a list of generic types

Question

If I have

template <typename T> struct A;
template <typename T> struct B;
template <typename T> struct C;
template <typename T> struct D;

what is the most compact way of testing if some candidate type X is one of them? I'm looking for something like

boost::enable_if< is_instantiation_of_any<X,A,B,C,D> >

but A,B,C and D are templates so I'm not sure how to construct the above.

Answer 1

With C++17 (or C++14 if you edit as per the comments), you can use boost::hana::any_of as a helper:

#include <iostream>
#include <boost/hana/any_of.hpp>
#include <boost/hana/tuple.hpp>
#include <type_traits>

namespace hana = boost::hana;

template <typename T> struct A {};
template <typename T> struct B {};
template <typename T> struct C {};
template <typename T> struct D {};
template <typename T> struct N {};

template<typename T>
bool fun(T) {      // overload for non-templated arguments (templated arguments
    return false;  // are a better match for the overload below)
}

template<typename T, template<typename> typename X>
bool fun(X<T>) { // overload for templated arguments
    auto constexpr pred = [](auto x){ // remove constexpr for < c++17
        return std::is_same_v<decltype(x), X<T>>;
        // return std::is_same<decltype(x), X<T>>::value; // for < c++17
    };
    return hana::any_of(hana::tuple<A<T>, B<T>, C<T>, D<T>>{}, pred);
};

int main() {
    A<int> a{};
    N<int> b{};
    int x = 3; fun(x);
    std::cout << fun(a) << fun(b) << fun(x) << std::endl; // prints 100
}

Probably you want to use std::decay_t before passing things to std::is_same_v.

Answer 2

Not sure if there exists a std::is_instantiation_of, though if all templates have same number of parameters, it is straightforward to implement it (if they don't it is more complicated). To check if a type is an instantiation of any of given templates you just need to fold it (requires C++17):

#include<iostream>
#include<type_traits>


template <typename T> struct A;
template <typename T> struct B;
template <typename T> struct C;
template <typename T> struct D;


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

template <typename A,template<typename> typename X>
struct is_instantiation_of<X<A>,X> : std::true_type {};

template <typename T,template<typename> typename...X>
struct is_instantiation_of_any {
    static const bool value = ( ... || is_instantiation_of<T,X>::value);
};

int main(){
    std::cout << is_instantiation_of< A<int>, A>::value;
    std::cout << is_instantiation_of< A<double>, B>::value;
    std::cout << is_instantiation_of_any< A<int>,A,B>::value;
}

Output:

101

---

To get a C++11 compliant solution, we can use this neat trick from one of Jarod42s answers:

template <bool ... Bs>
using meta_bool_and = std::is_same<std::integer_sequence<bool, true, Bs...>,
                                   std::integer_sequence<bool, Bs..., true>>;

Its rather clever, true,a,b,c and a,b,c,true are only the same when a, b and c are all true. std::integer_sequence is C++14, but all we need here is a type that has the bools as part of its definition:

namespace my {
    template <typename T,T ...t>
    struct integer_sequence {};
}

Using that we can rewrite the above to:

template <bool ... Bs>
using my_all = std::is_same<my::integer_sequence<bool, true, Bs...>,
                            my::integer_sequence<bool, Bs..., true>>;

And as "ANY(a,b,c,d,...)" is just "! ALL( !a, !b, !c, !d,...)" we can use:

template <bool ... Bs>
struct my_any { static constexpr bool value = ! my_all< ! Bs...>::value; };

to write is_instantiation_of_any in a C++11 friendly way:

template <typename T,template<typename> typename...X>
struct is_instantiation_of_any {
    static const bool value = my_any< is_instantiation_of<T,X>::value ...>::value;
};

Complete C++11 example