打印成员函数的地址

Question

struct Widget {
    void test() {}
};

int func() {}

int main() {
    std::cout << &Widget::test << std::endl;
    std::cout << Widget::test << std::endl;
    std::cout << func << std::endl;
    std::cout << &func << std::endl;
}

In this code only the second line of main function doesn't compile. The others print 1 . Why does it print 1 . Shouldn't print the address of function? And why second doesn't compile but first does?

Answer 1

Why does it print 1. Shouldn't print the address of function?

No. std::cout can print a void* , but there's no implicit conversion from function pointer types to void* (for neither regular function pointers nor pointer-to-member types). There's a conversion from function pointer types to bool though. That's what we end up with.

And why second doesn't compile but first does?

Because the standard requires you to use & to get the address of a member function.