C++ Question

[Alois Kraus]

Is it possible to call a specialized function of a base class ? I mean: class A { protected: virtual int f() { return -1; }; public: void DoFunc() { printf("\nValue: %d",f()); }; }; class B : public A { protected: virtual int f() { return 1; }; }; int main() { B Inst; Inst.DoFunc(); } pure virtual function does not help because the MFC macro IMPLEMENT_DYNCREATE makes the compiler think it get instantiated. Is it possible to get the class A to print 1 instead of -1 ?

[Alvaro Mendez]

This code should print 1. I went as far as to try it myself and it does. Regards, Alvaro

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Well done is better than well said. -- Benjamin Franklin (I actually prefer medium-well.)

[Not Active]

You mean? A.DoFunc() == 1 No, class A knows nothing about class B. B on the other hand knows about A because it is derived from it.

[Alois Kraus]

I didnt know of that. I always thought the function of the current namespace would be used. Just tested it. It works. cooool Thanks

[Alois Kraus]

I wrote Inst.DoFunc() in the example. This was not obvious to me. Inst.DoFunc() does indeed ignore f() member in class A and uses the derived f() from B.

[Jason Henderson]

If it is public or protected it will work.

Jason Henderson
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[retZ]

You should have a new public (not private or protected) member function in class B, inside which you would call A's DoFunc() function... There are no failures; there are only extended learning opportunities.

[Alvaro Mendez]

No problem. That's what the virtual keyword is for, in front of your f() function. I suggest you read up on it. Regards, Alvaro

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Well done is better than well said. -- Benjamin Franklin (I actually prefer medium-well.)

[Gary R Wheeler]

If you want to guarantee that the base class version will be called, rather than a virtual one from a derived class, you can do the following:

class A
{
protected:
virtual int f() { return -1; };

public:
void DoFunc() { printf("\nValue: %d",A::f()); };

};

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Software Zen: delete this;