Simultaneously overriding multiple member functions

[Mark P]

#include <iostream>
using namespace std;

struct Base
{
virtual int foo () {return 1;}
virtual int foo (int i) {return 2;}

virtual ~Base () {}
};

struct Derived : Base
{
virtual int foo (int i = 3) {return i;}
};

int main ()
{
Derived* d1 = new Derived;
Base* d2 = new Derived;
cout << d1->foo() << " " << d1->foo(4) << endl;
cout << d2->foo() << " " << d2->foo(4) << endl;
delete d2;
delete d1;
}

I was a little bit surprised to see the output of this short program:

3 4
1 4

Specifically, the '1' was unexpected. What rule (or rules) make it so
that d2->foo() calls Base::foo() rather than Derived::foo() with a
default argument?

Thanks,
Mark

[AnalogFile]

Re: Simultaneously overriding multiple member functions

Mark P wrote:[color=blue]
> #include <iostream>
> using namespace std;
>
> struct Base
> {
> virtual int foo () {return 1;}
> virtual int foo (int i) {return 2;}
>
> virtual ~Base () {}
> };
>
> struct Derived : Base
> {
> virtual int foo (int i = 3) {return i;}[/color]

This overrides Base::foo(int) and also defines a default argument for
it. The default argument will be in effect only if you call the method
from a reference to Derived.
It also hides Base::foo(). It does NOT override it. Base::foo(), being
hidden, cannot be called from a Derived object, unless you explicitelly
qualify the call.
[color=blue]
> };
>
> int main ()
> {
> Derived* d1 = new Derived;
> Base* d2 = new Derived;
> cout << d1->foo() << " " << d1->foo(4) << endl;[/color]

You are calling throu a Derived. The only visible member is
Derived::foo(in t). It has a default argument. Both calls are to that
method, the first one will use the default.
[color=blue]
> cout << d2->foo() << " " << d2->foo(4) << endl;[/color]

You are calling throu a Base (with dynamic type Derived). Both methods
are visible. The default value of Derived::foo(in t) is not in effect.
Base::foo() is not overridden. Base::foo(int) is overridden.

The first call is to Base::foo(), which is not overridden.

The only visible member is Derived::foo(in t). It has a default argument.
Both calls are to that method, the first one will use the default.
[color=blue]
> I was a little bit surprised to see the output of this short program:
>
> 3 4
> 1 4[/color]

We aren't ;-)
[color=blue]
>
> Specifically, the '1' was unexpected. What rule (or rules) make it so
> that d2->foo() calls Base::foo() rather than Derived::foo() with a
> default argument?[/color]

See above.
Also consider (note: I'm writing straight in the news client, forgive
any typing error):

/** ****** **/

struct Base {
virtual void foo();
virtual void foo(int);
};
struct Derived {
virtual foo();
};
.....
Derived d;
d.foo(2); // ERROR


the Base::foo(int) is hidden. If you want to call it you need:

d.Base::foo(2);



/** ****** **/

struct Base {
virtual void foo(int);
};
struct Derived {
virtual foo(int i=5);
};
.....
Base *p=new Derived;
p->foo(); // ERROR

the default is in effect for a call through Derived, but we are calling
through Base. And this is somewhat obvious: what would it be if you also
had:

struct Descendant: public Derived {
virtual foo(int i=7);
};

[AnalogFile]

Re: Simultaneously overriding multiple member functions

AnalogFile wrote:

Sorry. Some cut&paste mixup.
The previous post is incorrect. This part:
[color=blue][color=green]
>> cout << d2->foo() << " " << d2->foo(4) << endl;[/color]
>
> You are calling throu a Base (with dynamic type Derived). Both methods
> are visible. The default value of Derived::foo(in t) is not in effect.
> Base::foo() is not overridden. Base::foo(int) is overridden.
>
> The first call is to Base::foo(), which is not overridden.
>
> The only visible member is Derived::foo(in t). It has a default argument.
> Both calls are to that method, the first one will use the default.
>[/color]

Should have been:

[color=blue]
> cout << d2->foo() << " " << d2->foo(4) << endl;[/color]

You are calling through a Base (with dynamic type Derived). Both methods
are visible. The default value of Derived::foo(in t) is not in effect.
Base::foo() is not overridden. Base::foo(int) is overridden.

The first call is to Base::foo(), which is not overridden and is
therefore executed.
The second call is to Base::foo(int) which is overridden by
Derived::foo(in t), the latter is therefore executed.

[Mark P]

Re: Simultaneously overriding multiple member functions

AnalogFile wrote:[color=blue]
> Mark P wrote:[color=green]
>> #include <iostream>
>> using namespace std;
>>
>> struct Base
>> {
>> virtual int foo () {return 1;}
>> virtual int foo (int i) {return 2;}
>>
>> virtual ~Base () {}
>> };
>>
>> struct Derived : Base
>> {
>> virtual int foo (int i = 3) {return i;}[/color]
>
> This overrides Base::foo(int) and also defines a default argument for
> it. The default argument will be in effect only if you call the method
> from a reference to Derived.
> It also hides Base::foo(). It does NOT override it. Base::foo(), being
> hidden, cannot be called from a Derived object, unless you explicitelly
> qualify the call.
>[color=green]
>> };
>>
>> int main ()
>> {
>> Derived* d1 = new Derived;
>> Base* d2 = new Derived;
>> cout << d1->foo() << " " << d1->foo(4) << endl;[/color]
>
> You are calling throu a Derived. The only visible member is
> Derived::foo(in t). It has a default argument. Both calls are to that
> method, the first one will use the default.
>[color=green]
>> cout << d2->foo() << " " << d2->foo(4) << endl;[/color]
>
> You are calling throu a Base (with dynamic type Derived). Both methods
> are visible. The default value of Derived::foo(in t) is not in effect.
> Base::foo() is not overridden. Base::foo(int) is overridden.
>
> The first call is to Base::foo(), which is not overridden.
>
> The only visible member is Derived::foo(in t). It has a default argument.
> Both calls are to that method, the first one will use the default.
>[color=green]
>> I was a little bit surprised to see the output of this short program:
>>
>> 3 4
>> 1 4[/color]
>
> We aren't ;-)
>[color=green]
>>
>> Specifically, the '1' was unexpected. What rule (or rules) make it so
>> that d2->foo() calls Base::foo() rather than Derived::foo() with a
>> default argument?[/color]
>
> See above.
> Also consider (note: I'm writing straight in the news client, forgive
> any typing error):
>
> /** ****** **/
>
> struct Base {
> virtual void foo();
> virtual void foo(int);
> };
> struct Derived {
> virtual foo();
> };
> ....
> Derived d;
> d.foo(2); // ERROR
>
>
> the Base::foo(int) is hidden. If you want to call it you need:
>
> d.Base::foo(2);
>
>
>
> /** ****** **/
>
> struct Base {
> virtual void foo(int);
> };
> struct Derived {
> virtual foo(int i=5);
> };
> ....
> Base *p=new Derived;
> p->foo(); // ERROR
>
> the default is in effect for a call through Derived, but we are calling
> through Base. And this is somewhat obvious: what would it be if you also
> had:
>
> struct Descendant: public Derived {
> virtual foo(int i=7);
> };[/color]

Very clear and helpful! Thanks a lot.

Mark