编译时检查具有相同签名的构造函数

Question

是否可以在编译时检查属于两个不同类的两个构造函数是否具有相同的签名？ 如果可以，如何实现？

例子：

struct A
{
    A(int){}
};

struct B
{
    B(int){}
};

int main()
{
    static_assert(std::same_signature< A::A , B::B >::value, "A and B must have the same constructor parameters");

    return 0;
}

Answer 1

是否可以在编译时检查属于两个不同类的两个构造函数是否具有相同的签名？

不完全是你想要的，但你可以检查class A和class B是否可以 使用这种构造从相同的类型构造CheckConstructable<A, B, types...>::value，c++11：

#include <utility>
#include <string>
#include <type_traits>
#include <iostream>

struct A { A(int){} };

struct B { B(int){} B(std::string) {} };

struct C { C(std::string) {} };

template<class A, class B, typename... Types>
struct CheckConstructable;

template<class A, class B>
struct CheckConstructable<A, B> {
    static constexpr bool value = false;
};

template<class A, class B, typename T1, typename... Types>
struct CheckConstructable<A, B, T1, Types...> {
    static constexpr bool cur_type_ok = std::is_constructible<A, T1>::value && std::is_constructible<B, T1>::value;
    static constexpr bool value = cur_type_ok || CheckConstructable<A, B, Types...>::value;
};

int main()
{
    std::cout << "Have the same: " << (CheckConstructable<A, B, int, std::string>::value ? "yes" : "no") << "\n";
    std::cout << "Have the same: " << (CheckConstructable<A, B, std::string>::value ? "yes" : "no") << "\n";
    std::cout << "Have the same: " << (CheckConstructable<A, C, std::string>::value ? "yes" : "no") << "\n";
    std::cout << "Have the same: " << (CheckConstructable<B, C, std::string, int>::value ? "yes" : "no") << "\n";
}