VIrtual base function called in base class, C++

Question

I have a class:

class B
{
public:
  FuncCallingFoo();

protected:
   virtual Foo (const arg argument);
}:

Now, function Foo is used within the body of class B, i.e. it gets called somewhere in the definition of FuncCallingFoo.

I also have class A, which inherits from B and has it's own implementation of Foo:

class A:B
{
...

protected:
   Foo (const arg argument);
};

Let's say I create an instance of class A:

A a;

Furthermore, assume I do the following call:

a.FuncCallingFoo();

will this result in the implementation of Foo from class A being called?

That sounds like something you could easily try yourself, doesn't it? I'm not sure where the problem is — UnholySheep, Sep 14 '18 at 08:38
You could have tried it in less time than it took you to go here, write the question, and wait for an answer. — molbdnilo, Sep 14 '18 at 08:39
Indeed, it takes 2 mins to try this out: https://onlinegdb.com/S1vfUgtd7 — Frederik De Ruyck, Sep 14 '18 at 08:49

score 1 · Accepted Answer · answered Sep 14 '18 at 08:54

1

No, because you have declared Foo as protected in class A and B and you have not requested for public inheritance: class A:B (Yes, this is still valid C++).

Actually your program won't compile (after fixing the default return values and argument errors) due to this issue.

See: Difference between private, public, and protected inheritance

answered Sep 14 '18 at 08:54

Ferenc Deak

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Aha, I see, but by changing to class A: public B, that should solve the issue, I guess? – Wballer3 Sep 14 '18 at 09:15
Well, in that case yes, `A::foo` will be called – Ferenc Deak Sep 14 '18 at 09:31

VIrtual base function called in base class, C++

1 Answers1