Thread: Virtual functions called by non-vitual?

Ok, this is my code (well, not yet but if it works then yes it is ):

Code:

class Upgrade
{
public:
    bool collide();
protected:
    virtual void upgrade();
};

class Shields : public Upgrade
{
protected:
    void upgrade();
};

void Upgrade::upgrade()
{}

bool Upgrade::collide()
{
    if(/*collision*/)
    {
        //disappear (I'll figure it out later)

        upgrade();   // <---

        return true;
    }
    return false;
}

void Shields::upgrade()
{
    stuff();
}

If I then call collide(), and there is a collision, will Shields::upgrade() be called or will Upgrade::upgrade() be called?

Hmm, actually I thought that I needed a pointer to make it use the Shields version... but in Upgrade::upgrade(), it doesn't use a pointer. Does that matter?...

And why do I need the "= 0" on the declaration?

Oh, ok... I'll just grab a dictionary

You'd include this if you wanted to prevent an Upgrade object from be created and to force all directly dervied classes to implement ::upgrade

That means "if you wanted to prevent an Upgrade object from being created and if you wanted to force all directly derived classes to implement ::upgrade"? Or "if you... and if you didn't want to force all (...)"?

He is right.

Code:

class Upgrade
{
public:
    bool collide();
protected:
    virtual void upgrade();
};

void Upgrade::upgrade(){cout << "class Upgrade calls upgrade" << endl;}

bool Upgrade::collide()
{
    if(1 == 1)
    {
        //disappear (I'll figure it out later)

        upgrade();   // <---

        return true;
    }
    return false;
}




class Shields : public Upgrade
{
protected:
    void upgrade();
};




void Shields::upgrade()
{
 cout <<  "class Shields calls upgrade" << endl;
}




int main(int argc, char *argv[])
{
 Shields s;
 s.upgrade();
 s.collide();
 getch();

 return 0;
}

As you can see, after running the program, both s.upgrade() and s.collide() called the Sheilds version of 'upgrade()'. Note though, you could ensure that in 'collide()' that the base version of upgrade() would be called if you defined:

Code:

bool Upgrade::collide()
{
    if(1 == 1)
    {
        //disappear (I'll figure it out later)

        Upgrade::upgrade();   // <---like so.

        return true;
    }
    return false;
}

Ok, so if i do this:

Code:

Shields s;
if(s.collide())
    //(do something)

then Upgrade::collide() will be called, and it will in turn call Shields::upgrade()? (sorry, I'm slow )

Yes.

Ok, thanks guys!

Just for reference here is a little terminology. When you define a function in your base class with the " = 0" it is known as a 'pure virtual function' , and any class containing pure virtual functions are called 'abstract base classes'.

Okee... and are 'abstract base classes' good?

Oh n/m, i'll look it up on MSDN