Re: inheritance - method resolution

From:
Pete Becker <pete@versatilecoding.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 7 Apr 2011 15:15:02 -0400
Message-ID:
<201104071515026212-pete@versatilecodingcom>
On 2011-04-07 15:10:59 -0400, Leigh Johnston said:

On 07/04/2011 20:08, Pete Becker wrote:

On 2011-04-07 13:12:10 -0400, Christopher said:

Where is the rule that explains why this will not compile? I've always
expected this to work, but it would appear that I haven't run into
this problem yet.

To resolve the problem, do I really need to override every single
method from the Base with the same name as the specific method I am
interested in overriding? I have a good 20 of them in production code.

A simple test case to reproduce what I am experiencing in more
complicated code:

class Base
{
public:
virtual void Foo()
{
}

void Foo(int x)
{

}
};

class Derived : public Base
{
public:
void Foo()
{
}
};

int main()
{

Derived * blah = new Derived();
blah->Foo(5);
delete blah;

return 0;
}


Overloading applies to names defined in the same scope. Base defines two
versions of Foo, so when you have an object of type Base you can use
overloading to select which function to call. Derived, however, defines
one version of Foo, so there's nothing to overload. As others have said,
adding a using directive in Derived tells the compiler to pretend that
all of the Foos defined in Base were also defined in Derived; now you
have two versions of Foo in Derived, and you can use overloading to pick
the one to call.


Almost correct;


And certainly more useful than your condescending responses.

 it is actually called a "using declaration" not a "using directive"
which is something different.


Thank you for your input.

--
  Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

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