类成员函数内的虚拟函数调用

Question

I understand the basic concept of virtual function and vtable, but in the following example, I don't understand why cA(); prints out

parent A
child

but without the virtual keyword for Parent::func(), it prints out

parent A
parent

Would you let me know the reason in detail? It would be great to explain with v table, memory (heap, stack), etc..

Thanks.

#include <iostream> 

template <class TYPE> class Parent
{
public:
    Parent() {};
    ~Parent() {};
    virtual void func() { std::cout << "parent" << std::endl; };

    void A() {
        std::cout << "parent A" << std::endl;
        func();
    }

};

template <class TYPE> class Child : public Parent <TYPE>
{
public:
    Child() {};
    ~Child() {};

    void func() { std::cout << "child" << std::endl; };
};

void main()
{
    Child<int> c;
    c.A();
}

Answer 1

The virtual key word specifies that the function can be redefined in a derived class, while preserving its calling properties though references. This is basically the trigger for polymorphic behavior. If the function is declared virtual and it is redefined in a derived class then the vtable is utilized to select the appropriate version of the function unless a specific namespace is specified. For example Parent::func(). Despite having the same name, without the key word virtual the two functions you named func() are completely different. There is no reference available to the base class that can access the derived class's version of the function. It uses the only version func() it knows about, which is the one defined in the base class.

Answer 2

@Pemdas gave a nice explanation. Here is my try.

cA() tells compiler "I am gonna call the non virtual function A defined in my parent class". This translates to Parent::A(&c)

The A method in class Parent translates to "get the vtable of object &c, grab the function pointer in first row, and call it". Since c reimplements the function func , the function pointer would be c 's implementation of function func . That would be what you see.