About abstract classes and Inheritance

[Barzo]

Hi,

I have a newbie doubt. In the following code, why BaseAImpl is
abstract? Deriving it from BaseImpl the IBase::foo() is not defined?




#include<iostream>

using namespace std;

class IBase
{
public:
virtual void foo() = 0;
};

class IBaseA : public IBase
{
public:
virtual void foo_bar_A() = 0;
};

class IBaseB : public IBase
{
public:
virtual void foo_bar_B() = 0;
};


class BaseImpl : IBase
{
public:
void foo(){cout << "BaseImpl::foo" << endl;};
};

class BaseAImpl : public IBaseA, public BaseImpl
{
public:
void foo_bar_A(){cout << "BaseAImpl::foo_bar_A" << endl;};
};

class BaseBImpl : public IBaseB, public BaseImpl
{
public:
void foo_bar_B(){cout << "BaseBImpl::foo_bar_B" << endl;};
};



int _tmain(int argc, _TCHAR* argv[])
{
IBaseA* A = new BaseAImpl;
A->foo_bar_A();

IBase* B = (IBase*)A;
B->foo();

delete A;

return 0;
}

 

[Victor Bazarov]

I have a newbie doubt.

A doubt? A doubt you have when you are uncertain of some belief or
opinion, or when you lack confidence in your abilities.

You actually just have *two questions*.

> In the following code, why BaseAImpl is
abstract? Deriving it from BaseImpl the IBase::foo() is not defined?

'BaseAImpl' has two objects of type 'IBase'. One inside 'BaseImpl'
subobject, and the other inside the 'IBaseA' subobject. The former
defines the final overrider for 'IBase::foo', the latter does not.
That's why it's abstract. And, no, one base class' final overrider does
NOT become the final overrider for another base class' virtual function.

[..]

V

 

[Barzo]

A doubt?  A doubt you have when you are uncertain of some belief or
opinion, or when you lack confidence in your abilities.

You actually just have *two questions*.

Ok, I have two questions..

 > In the following code, why BaseAImpl is


'BaseAImpl' has two objects of type 'IBase'.  One inside 'BaseImpl'
subobject, and the other inside the 'IBaseA' subobject.  The former
defines the final overrider for 'IBase::foo', the latter does not.
That's why it's abstract.  And, no, one base class' final overrider does
NOT become the final overrider for another base class' virtual function.

Thanks, now I understand.

I solved using 'virtual' Inheritance:

class IBase {...};

class IBaseA : virtual public IBase {...};

class BaseImpl : virtual public IBase {...};

class BaseAImpl : public IBaseA, public BaseImpl {...};

In this manner only one IBase object will be created, right?

Daniele.

 

[Victor Bazarov]

Ok, I have two questions..


Thanks, now I understand.

I solved using 'virtual' Inheritance:

class IBase {...};

class IBaseA : virtual public IBase {...};

class BaseImpl : virtual public IBase {...};

class BaseAImpl : public IBaseA, public BaseImpl {...};

In this manner only one IBase object will be created, right?

Right.

V