Can I Determine if a Class Is a Specialization of a Class Template?

Can I Confirm Class Template Specialization?

In software development, we often need to determine if a given class is specialized due to class templates. Consider the following scenario:

Problem:
Given a class template like

template <class T>
struct A {};

Is it possible to ascertain whether CompareT is an instance of A<> for any type *? For example, in the below code:

template<class CompareT>
void compare(){
   // is this A ?
   cout << is_same< A<*> , CompareT >::value;     // A<> ????
}

int main(){
  compare< A<int> >();
}

In this use case, A should align with A<>, resulting in an output of 1.

Solution:

The below code allows you to verify if a class is a specialized version of a template:

template <class T, template <class...> class Template>
struct is_specialization : std::false_type {};

template <template <class...> class Template, class... Args>
struct is_specialization<Template<Args...>, Template> : std::true_type {};

static_assert(is_specialization<std::vector<int>, std::vector>{}, "");
static_assert(!is_specialization<std::vector<int>, std::list>{}, "");

By invoking is_specialization, you can identify if a class is a template specialization, granting you finer control over your code's structure and behavior.

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