I want to create a class specialization that has the same implementation if it gets passed any std::variant or any boost::variant. I tried to play around with std::enable_if, std::disjunction and std::is_same but I couldn't make it compile. Here is a code sample to show what I want to achieve.
#include <variant>
#include <iostream>
#include <boost/variant.hpp>
template <typename T>
struct TypeChecker;
template <typename T>
struct TypeChecker<T>
{
void operator()()
{
std::cout << "I am other type\n";
}
}
template <typename ... Ts> // I want to be able to capture Ts... in TypeChecker scope
struct TypeChecker<std::variant<Ts...> or boost::variant<Ts...>> // what to insert here?
{
void operator()()
{
std::cout << "I am either std::variant or boost::variant\n";
}
}
int main()
{
TypeChecker<std::variant<int, float>>{}();
TypeChecker<boost::variant<int, float>>{}();
TypeChecker<int>{}();
}
Expected result:
I am either std::variant or boost::variant
I am either std::variant or boost::variant
I am other type
You can use a template template parameter e.g. like this
template <typename T>
struct TypeChecker {
void operator()() {
std::cout << "I am other type\n";
}
};
template<typename ... Ts, template<typename...> typename V>
requires std::same_as<V<Ts...>, std::variant<Ts...>> ||
std::same_as<V<Ts...>, boost::variant<Ts...>>
struct TypeChecker<V<Ts...>>
{
void operator()()
{
std::cout << "I am either std::variant or boost::variant\n";
}
};
Note that this uses C++20 constraints. If you can't use C++20 you can use std::enable_if
instead like e.g. this:
template<typename ... Ts, template<typename...> typename V>
struct TypeChecker<V<Ts...>> : std::enable_if_t<
std::is_same_v<V<Ts...>, std::variant<Ts...>> ||
std::is_same_v<V<Ts...>, boost::variant<Ts...>>, std::true_type>
{
void operator()()
{
std::cout << "I am either std::variant or boost::variant\n";
}
};
You can see it live on godbolt.