asked on

Derived class with default function parameters: needs explanation please !

Ah; hello !

This is a hardcore C++ question that I am hoping can be resolved by someone who can give me a solid explanation.

Please peruse this code:

class A
{
public:
A() {}
virtual int GetVal(int a = 1) { return a; }
};

class B : public A
{
public:
B() {}
virtual int GetVal(int b = 2) { return b; }
}

main()
{
A *x = new B();
int r = x->GetVal();
}

Confusion: r is 1. Now I know this, but would like solid explanation and clarification as to *why*.

Firstly we are dealing with pointers, so that allows run time binding, right ? And we are creating a B object but assigning it to an A pointer. So why is the output not 2 ? Arghhhh !

TIA

jkr

The default parameter's value is deducted from the pointer type (where else should the compiler get that info from?) - consider the following:

class I
{
public:
virtual int GetVal(int a = 0) = 0;
};

class A : public I
{
public:
A() {}
virtual int GetVal(int a = 1) { return a; }
};

class B : public A
{
public:
B() {}
virtual int GetVal(int b = 2) { return b; }
}

main()
{
I *x = new B();
int r = x->GetVal();
}

'r' will be (in fact is) 0...

mrwad99

ASKER

So you are saying that as a general rule, default parameters are resolved at compile time always ?

I have just tried the same without pointers and the default value is taken from the type being assigned *to*, in this case I.

I am still a little confused however as to why it is taking the default from I when it can clearly see that I want to make an B object...

ASKER CERTIFIED SOLUTION

jkr

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mrwad99

ASKER

OK. I guess I need to re-read about virtual functions and late binding then.

Thanks.

jkr

No, in general, you would be right. I't's just that 'default parameter' thingy that messes things up here.