类声明外部的SFINAE模板实现

#include <iostream>
#include <type_traits>

// 检测类型T是否有名为foo的成员函数
template <typename T, typename = void>
struct has_foo : std::false_type {};

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

// 根据T是否有foo成员函数来选择不同的实现
template <typename T>
void call_foo(T& obj, std::true_type) {
    obj.foo();
}

template <typename T>
void call_foo(T& obj, std::false_type) {
    std::cout << "Type does not have foo() member function." << std::endl;
}

template <typename T>
void call_foo(T& obj) {
    call_foo(obj, has_foo<T>{});
}

struct A {
    void foo() { std::cout << "A::foo()" << std::endl; }
};

struct B {};

int main() {
    A a;
    B b;
    call_foo(a); // 输出: A::foo()
    call_foo(b); // 输出: Type does not have foo() member function.
    return 0;
}