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In my program I am using two classes foo1, the parent class, and foo2
the child class!
foo2 has overloaded inserter and extractor operator, but so does foo1!
How can I access foo1 inserter or extractor?


Thanks

On 2008-06-01 03:07, Victor Bazarov wrote:
> ArbolOne wrote:
>> In my program I am using two classes foo1, the parent class, and foo2
>> the child class!
>> foo2 has overloaded inserter and extractor operator, but so does foo1!
>> How can I access foo1 inserter or extractor?
>
> Post your code. It is easier when there is substance to the discussion.
>
> Meanwhile, here is the example:
>
> struct foo {
> void foobar();
> };
>
> struct bar : foo {
> void foobar();
> };
>
> int main() {
> bar b;
> b.foobar(); // calls bar::foobar
> b.foo::foobar(); // call foo::foobar
> }

I would call it bad style to explicitly call a parent's implemenetation
of a function (and bad design if you have to), unless it is done in the
child's implementation of the same function:

#include<iostream>

struct foo1 {
virtual void foobar() {
std::cout << "foo1\n";
}
};

struct foo2 : public foo1 {
virtual void foobar() {
std::cout << "foo2\n";
foo1::foobar(); // Call foo1's foobar()
}
};

int main() {
foo1* f = new foo1();
f->foobar();
delete f;
f = new foo2();
f->foobar();
}

--
Erik Wikström

On Jun 1, 3:57 am, Erik Wikström <Erik-wikst...@telia.com> wrote:
> On 2008-06-01 03:07, Victor Bazarov wrote:
>
>
>
> > ArbolOne wrote:
> >> In my program I am using two classes foo1, the parent class, and foo2
> >> the child class!
> >> foo2 has overloaded inserter and extractor operator, but so does foo1!
> >> How can I access foo1 inserter or extractor?
>
> > Post your code. It is easier when there is substance to the discussion.
>
> > Meanwhile, here is the example:
>
> > struct foo {
> > void foobar();
> > };
>
> > struct bar : foo {
> > void foobar();
> > };
>
> > int main() {
> > bar b;
> > b.foobar(); // calls bar::foobar
> > b.foo::foobar(); // call foo::foobar
> > }
>
> I would call it bad style to explicitly call a parent's implemenetation
> of a function (and bad design if you have to), unless it is done in the
> child's implementation of the same function:
>
> #include<iostream>
>
> struct foo1 {
> virtual void foobar() {
> std::cout << "foo1\n";
> }
>
> };
>
> struct foo2 : public foo1 {
> virtual void foobar() {
> std::cout << "foo2\n";
> foo1::foobar(); // Call foo1's foobar()
> }
>
> };
>
> int main() {
> foo1* f = new foo1();
> f->foobar();
> delete f;
> f = new foo2();
> f->foobar();
>
> }
>
> --
> Erik Wikström
I got it now!
Thanks.