如果我尝试在 c++ 中打印类的非静态成员函数的地址，为什么会创建 object？

Question

I found one interesting behavior while I was just checking non-static member variable's offsets

It looks like if I try to print non-static member function's address in pointer
It just creates a new object for the class which has the member function

Is this a normal behavior or undefined behavior that I can just ignore?

for instance if I try below code

#include <iostream>

typedef void(*FP)();

class AAA {
public:
    int var1;
    int var2;

    void foo() {
        std::cout << &var1 << std::endl;
        std::cout << &var2 << std::endl;
    }
};

int main() {
    FP fp = reinterpret_cast<FP>(&AAA::foo);
    std::cout << fp << std::endl;
    fp();
}

Its result is like below

$ ./a.out
0x56001ea8d940
0x7fe36c93c760
0x7fe36c93c764

So I removed cout statement then it just prints member's offset like I expected

#include <iostream>

typedef void(*FP)();

class AAA {
public:
    int var1;
    int var2;

    void foo() {
        std::cout << &var1 << std::endl;
        std::cout << &var2 << std::endl;
    }
};

int main() {
    FP fp = reinterpret_cast<FP>(&AAA::foo);
    fp();
}

and its result is like below

$ g++ main.cpp (with some warnings anyway)
$ ./a.out
0x1
0x5

Answer 1

The weird behaviour you're seeing is undefined behaviour caused by the following line:

FP fp = reinterpret_cast<FP>(&AAA::foo);

Even without calling fp this is already undefined behaviour! Only the listed conversions are valid for reinterpret_cast ( [expr.reinterpret.cast] ), and the one you're using (member function pointer to function pointer) is not one of them.

It's a common misunderstanding that reinterpret_cast is an “anything goes” cast that allows circumventing type safety in any way imaginable. According to the standard, that's not the case; only a handful of specific conversions are valid. All others are not (but do not require diagnosis, which leads to them being used anyway).