Base class pointer assigned to derived class

Question

I created base class pointer and assigned new derived class to it.

 class Base
{ 
public: 
    virtual void f( int ); 
    virtual void f( double ); 
    virtual void g( int i = 10 ); 
}; 

void Base::f( int )
{ 
    cout << "Base::f(int)" << endl;
} 

void Base::f( double )
{ 
    cout << "Base::f(double)" << endl;
} 

void Base::g( int i )
{
    cout << i << endl;
} 

class Derived: public Base
{ 
public: 
    void f( complex<double> ); 
    void g( int i = 20 ); 
};

void Derived::f( complex<double> )
{ 
    cout << "Derived::f(complex)" << endl;
} 

void Derived::g( int i )
{ 
    cout << "Derived::g() " << i << endl;
} 

void main()
{ 
    Base b; 
    Derived d; 
    Base* pb = new Derived; 
    b.f(1.0); 
    d.f(1.0); 
    pb->f(1.0); 
    b->g();
    d->g();
    pb->g(); 
    delete pb; 
}

Result is:

Base::f(double) 
Derived::f(complex) 
Base::f(double) 
10 
Derived::g() 20 
Derived::g() 10

Results for b and d are expected. pb->f(1.0) calls Base function f(double), but pb->g() seems to call Derived class function g but using parameter i=10 from Base class. Why?

Why? Because you're using the Base interface statically, the dispatch to Derived::pb happens only at runtime. — Kerrek SB
– Kerrek SB, Commented Mar 12, 2017 at 21:22
delete pb; causes undefined behaviour, since Base does not have a virtual destructor — M.M
– M.M, Commented Mar 13, 2017 at 2:18

Community · Accepted Answer · 2017-05-23 11:53:49Z

2

Kerrek SB said it all in the comment, I will detail it a bit.

As in this answer, the mechanism of resolving default function parameters occurs at compile time. When the compiler sees pb->g();, knowing that pb is a pointer of type Base*, it consults the declaration of Base::g, that is

virtual void g( int i = 10);

It concludes:

1- The missing parameter i should be set to 10. This is a compile-time decision.

2- The method is declared virtual, which means that the actually invoked method depends on what pb will be pointing-to at run time. The compiler sets the mechanism to dispatch the call to g at runtime (usually through an indirection via the vtable). Since at runtime it happens that pb is actually pointing to an object of type Derived, the method Derived::g is called.

edited May 23, 2017 at 11:53

CommunityBot

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answered Mar 12, 2017 at 22:35

A.S.H

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