Implement is by yourself_ convertible

c++ Provided in the standard library std::is_convertible, Used to determine whether it can be implicitly converted from one type to another , Here's an example ：

class Ax
{
    

};

class Bx:public Ax
{
    
public:
};

class Cx
{
    

};

int main()
{
    
	cout << std::is_convertible<Bx, Ax>::value << endl;
	cout << std::is_convertible<Ax, Bx>::value << endl;
	cout << std::is_convertible<float, int>::value << endl;
	cout << std::is_convertible<int, float>::value << endl;
	system("pause");
	return 0;
}

result ：

After understanding its function , We try to achieve it by ourselves .

template<typename FROM,typename TO>
struct ISCONVERTIBLE
{
    
private:
	static void testfunc(TO);

	template<typename = decltype(testfunc(std::declval<FROM>()))>
	static std::true_type test(void *);

	static std::false_type test(...);
public:
	static  constexpr bool value = decltype(test(nullptr))::value;
};

The above is the implementation code , Use overloaded functions to realize , Return respectively std::true_type,std::false_type; If FROM Can be converted into TO type , Then it will match std::true_type Of test() Member functions , Otherwise, it will match std::false_type Of test Member functions . Be careful : If FROM The type can be successfully converted to TO type , Then through the decltype infer testfunc() The return type of is valid ,test The function will return std::true_type, Otherwise, the compiler will match and return std::false_type Member function of . Let's do a test ：