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Home/ Questions/Q 7158681
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T13:06:23+00:00 2026-05-28T13:06:23+00:00

My Code: #include <iostream> using namespace std; class A { public: virtual void print(void)

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My Code:

#include <iostream>

using namespace std;

class A
{
public:
    virtual void print(void) { cout << "I am base class" << endl; }
};

class B : public A
{
public:
    void print(void) { cout << "I am class B" << endl; }
};

void mainprint(A *a)
{
    (*a).print();
}

int main()
{
    A a;
    B b;

    B *bp;
    A *ap;

    ap = &b;

    a.print();
    b.print();
    (*ap).print();

    bp = new B();

    mainprint((A *)bp);

    delete bp;

    return 0;
}

Output:

I am base class
I am class B
I am class B
I am class B

I have casted the pointer(bp) to class A inside the function call, but it still calls the derived class print!!!

Can someone shed some light on this for me.

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  1. Editorial Team
    Editorial Team
    2026-05-28T13:06:24+00:00Added an answer on May 28, 2026 at 1:06 pm

    I have casted the pointer(bp) to class A inside the function call, but it still calls the base class print!!!

    I assume you mean “calls the derived class print”, since that’s what happened.

    That’s the whole point of virtual functions; the final override associated with the actual type of the object (i.e. the “dynamic type”) is chosen, whatever the type of the reference or pointer used to call the function (i.e. the “static type”). So B::print is chosen, because bp still points to an instance of B.

    If you want to force a call to A::print, you could do:

    pb->A::print()
    

    or, if you don’t want polymorphic behaviour at all, remove the virtual specification.

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