在使用SFINAE的模板类之外定义函数？

#include <iostream>
#include <type_traits>

// 辅助结构体，用于检查类型是否有某个成员函数
template <typename T, typename = void>
struct has_member_function : std::false_type {};

template <typename T>
struct has_member_function<T, std::void_t<decltype(std::declval<T>().member_function())>> : std::true_type {};

// 模板类
template <typename T>
class MyClass {
public:
    void do_something() {
        if constexpr (has_member_function<T>::value) {
            std::cout << "Type has member function." << std::endl;
        } else {
            std::cout << "Type does not have member function." << std::endl;
        }
    }
};

// 在模板类之外定义函数
template <typename T>
void external_function(T& obj) {
    if constexpr (has_member_function<T>::value) {
        std::cout << "External function: Type has member function." << std::endl;
        obj.member_function();
    } else {
        std::cout << "External function: Type does not have member function." << std::endl;
    }
}

// 测试类
struct WithMemberFunction {
    void member_function() {
        std::cout << "Member function called." << std::endl;
    }
};

struct WithoutMemberFunction {};

int main() {
    MyClass<WithMemberFunction> obj1;
    obj1.do_something();

    MyClass<WithoutMemberFunction> obj2;
    obj2.do_something();

    WithMemberFunction wm;
    external_function(wm);

    WithoutMemberFunction wom;
    external_function(wom);

    return 0;
}