calling non-static member function without instantiation using `reinterpret_cast`

问题

As it's shown in the following code, I can call a non-static member function A::f without instantiating an object of the class. This is only possible when the function is not bound to any other member. For example, I cannot call A::g in a similar fashion.

It seems to me that the call to A::f as shown in the code below behaves like calling a static member function. Is such a conclusion correct? How can this behavior be justified?

#include <iostream>
using namespace std;

struct A {
    void f() { cout << "Hello World!"; }    
    void g() { cout << i; } 
    int i = 10;
};

int main() {
    auto fPtr = reinterpret_cast<void(*)()>(&A::f);
    (*fPtr)(); // OK

//  auto gPtr = reinterpret_cast<void(*)()>(&A::g); 
//  (*gPtr)(); // Error!
    return 0;
}

回答1:

Is such a conclusion correct? How can this behavior be justified?

It's undefined behavior.

Pretty useless to ask for how it should specifically behave. Pick one or more from the list:

²⁾Schroedingers cat is always found dead when you open the chest¹
Your fridge explodes
You see the behavior you get
Little demons are flying out from your nostrils
¹⁾You get time warped back to the point, when you put the living cat into the chest²
...
All of the above happens simultaneously

void f() { cout << "Hello World!"; }

Well, the fact that your code doesn't rely on any member variable data makes it likely, that any decent implementation works without "crashing".

Though, it's simply UB to call a class member function without a valid class instance, as mentioned above.
So you can't predict what happens, or should rely on your observations.

来源：https://stackoverflow.com/questions/37579850/calling-non-static-member-function-without-instantiation-using-reinterpret-cast

标签

c++

casting