C++ SFINAE在引用类型和转换为引用的参数之间选择重载？

#include <iostream>
#include <type_traits>

template <typename T>
typename std::enable_if<std::is_reference<T>::value>::type foo(T t) {
    std::cout << "Reference type" << std::endl;
}

template <typename T>
typename std::enable_if<!std::is_reference<T>::value &&
                        std::is_convertible<T, int&>::value>::type foo(T t) {
    std::cout << "Convertible to reference type" << std::endl;
}

int main() {
    int x = 42;
    int& ref = x;
    const int& const_ref = x;

    foo(ref);         // 输出：Reference type
    foo(const_ref);   // 输出：Reference type
    foo(42);          // 输出：Convertible to reference type

    return 0;
}