C++：通过返回类型进行重载解析？

Question

函数重载解析不考虑返回类型，但是我在Cppreference

上遇到了一段代码

SFINAE 不谈，这里的重载解析是如何发生的：

#include <iostream>
 
// This overload is added to the set of overloads if C is
// a class or reference-to-class type and F is a pointer to member function of C
template<class C, class F>
auto test(C c, F f) -> decltype((void)(c.*f)(), void())
{
    std::cout << "(1) Class/class reference overload called\n";
}
 
// This overload is added to the set of overloads if C is a
// pointer-to-class type and F is a pointer to member function of C
template<class C, class F>
auto test(C c, F f) -> decltype((void)((c->*f)()), void())
{
    std::cout << "(2) Pointer overload called\n";
}
 
// This overload is always in the set of overloads: ellipsis
// parameter has the lowest ranking for overload resolution
void test(...)
{
    std::cout << "(3) Catch-all overload called\n";
}
 
int main()
{
    struct X { void f() {} };
    X x;
    X& rx = x;
    test(x, &X::f);  // (1)
    test(rx, &X::f); // (1), creates a copy of x
    test(&x, &X::f); // (2)
    test(42, 1337);  // (3)
}

我已经查过了。这段代码有效。那么，这里是否使用尾随返回类型来进行重载解析？

Answer 1

重载解析不考虑返回类型

它们意味着，给定几个可行的重载，它们不会根据返回类型（是否与返回值的使用方式匹配，等等）消除歧义。

另一方面，当重载由于 SFINAE 错误而被拒绝时，错误是在返回类型中还是在其他地方并不重要。