Question on Virtual Methods

Question

IF both methods are declared as virtual, shouldn't both instances of Method1() that are called be the derived class's Method1()?

I am seeing BASE then DERIVED called each time. I am doing some review for an interview and I want to make sure I have this straight. xD

class BaseClass
{
public:
    virtual void Method1()  { cout << "Method 1 BASE" << endl; }
};

class DerClass: public BaseClass
{
public:
    virtual void Method1() { cout << "Method 1 DERVIED" << endl; }
};


DerClass myClass;
    ((BaseClass)myClass).Method1();
    myClass.Method1();

Method 1 BASE
Method 1 DERVIED

"IF both methods are declared as virtual," - The override is virtual anyway in the derived class (although people just prefer to add it there for clarity), so this is irrelevant to the question. — UncleBens, May 05 '10 at 21:11
The cast you're looking for would be `((BaseClass&)myClass).Method1();`, or a bit more C++: `static_cast(myClass).Method1();` — GManNickG, May 05 '10 at 21:14

score 14 · Answer 1 · answered May 05 '10 at 21:10

14

No, the "C-style" cast ((BaseClass)myClass) creates a temporary BaseClass object by slicing myClass. It's dynamic type is BaseClass, it isn't a DerClass at all so the Method1 being called is the base class method.

myClass.Method1() is a direct call. As myClass is an object, not a reference there is no virtual dispatch (there would be no need).

answered May 05 '10 at 21:10

CB Bailey

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Good call on the slice. It's always frames a good question to ask candidates to see if they have a clear distinction between C++ and say Java or C#. – Nathan Ernst May 06 '10 at 00:57

score 12 · Answer 2 · answered May 05 '10 at 21:08

12

No, because the virtual function mechanism only works if a function is called via a pointer or a reference. Otherwise the static type of the object is used to determine which function to call.

answered May 05 '10 at 21:08

1

Ah. Thanks everyone for the explanation! :D – bobber205 May 05 '10 at 21:19

Dima · Accepted Answer · 2010-05-05T21:54:15.707

6

What you are seeing here is called "slicing". Casting an object of the derived class to the base class "slices off" everything that is not in the base class.

In C++ virtual functions work correctly only for pointers or references. For your example to work right, you have to do the following:


DerClass myClass;
((BaseClass *) &myClass)->Method1();

Or you could do


BaseClass *pBase = new DerClass;
pBase->Method1();

edited May 05 '10 at 21:54

answered May 05 '10 at 21:49

Dima

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Dima: +1 for not forgetting to mention `In C++ virtual functions work correctly only for pointers or references` – nitin_cherian Jan 23 '12 at 09:55

score 5 · Answer 4 · answered May 05 '10 at 21:09

Because the cast ((BaseClass)myClass) slices the myClass object from DerClass to BaseClass, so only the BaseClass's implementation of Method1() is called.

For polymorphism to work properly, you must call the methods via pointers:

DerClass myClass; 
BaseClass* ptrToMyClass = &myClass;
ptrToMyClass->Method1(); // Calls the DerClass implementation of Method1()

or references:

DerClass myClass; 
BaseClass& refToMyClass = myClass;
refToMyClass.Method1();  // Calls the DerClass implementation of Method1()

score 0 · Answer 5 · answered Jan 29 '15 at 09:13

((BaseClass)myClass).Method1(); --> object slicing hence always base class method will get called. Real polymorphic behaviour is achieved through base class pointer which can contain any objects of derived classes. Hence to achieve what you want you need to do pass address of derived class object and typecase it to base class pointer. as below: ((BaseClass *) &myClass)->Method1();

Question on Virtual Methods

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