Why a derived object is not calling on the virtual function

Question

class Base{

public:
    virtual void func2() const{
        cout << "1" << endl;
    }

    virtual void func1(Base temp){
        cout << "Class 1" << endl;
        temp.func2();
    }

};
class Derived : public Base{
public:

    void func2() const{
        cout << "2" << endl;
    }
    void func1(Base temp){
        cout << "Class 2" << endl;
        temp.func2();
    }

};

int main(){
    Base one;
    Derived two;

    two.func1(two);
    cout << "------------\n";


    return 0;
}

Output is:

Class 2

1

So my question is: why when I'm giving func1(two) an object of Derived, it is not calling on virtual func2 that outputs 2. The compiler is always choosing func2() that is defined in the Base class.

It does, however, chooses func2 in the Derived class when you call on it, i.e. two.func2(); which outputs 2

try passing the argument to `func1` by reference, like so: `void func1(Base& temp)`. Also check @Wintermute's comment for more insight. — Tomislav Dyulgerov, Apr 07 '15 at 12:21

score 3 · Accepted Answer · answered Apr 07 '15 at 12:20

3

func1(Base temp)

That is copying your object, it is not your original Derived but a Base instance copy initialized from the derived.

func1(Base& temp) should work

answered Apr 07 '15 at 12:20

Mario Corchero

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score 1 · Answer 2 · answered Apr 07 '15 at 12:21

1

It's because you pass an argument by value: func1 param is Base temp, not Base &temp. So you construct a new, temporary object of class Base, so you call Base::func2() which prints 1.

answered Apr 07 '15 at 12:21

CiaPan

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score 1 · Answer 3 · answered Apr 07 '15 at 12:21

1

you should try with:

virtual void func1(Base& temp){
        cout << "Class 1" << endl;
        temp.func2();
    }

this will prevent copying the object, preserving the hierarchy.

answered Apr 07 '15 at 12:21

Felice Pollano

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Why a derived object is not calling on the virtual function

3 Answers3