héritage - C++: comment appeler une fonction de classe parent du dehors - Stack Overflow

I have:

class A{
    public:
        virtual void foo();
};

class B : public A{
    public:
        void foo();
};

B *ptr = new B();

I want to call A's foo() DIRECTLY using the 'ptr' pointer.

When I try

(A*)ptr->foo();

it still calls B's version of foo(). How do I call A's version instead?

Is this possible? What are the alternatives? Thank you.

When you name a function with the :: scope-resolution form, you call the named function, as though it were not virtual.

ptr->A::foo();

You need to make your functions public. You do this simply by making the following change:

class A{
    public:
        virtual void foo();
};

class B : public A{
    public:
        void foo();
};

When you don't do this, the functions are automatically private and inaccessible from the "outside".

I have:

class A{
    public:
        virtual void foo();
};

class B : public A{
    public:
        void foo();
};

B *ptr = new B();

I want to call A's foo() DIRECTLY using the 'ptr' pointer.

When I try

(A*)ptr->foo();

it still calls B's version of foo(). How do I call A's version instead?

Is this possible? What are the alternatives? Thank you.

When you name a function with the :: scope-resolution form, you call the named function, as though it were not virtual.

ptr->A::foo();

You need to make your functions public. You do this simply by making the following change:

class A{
    public:
        virtual void foo();
};

class B : public A{
    public:
        void foo();
};

When you don't do this, the functions are automatically private and inaccessible from the "outside".

Source