Re: How to do "derived" type extensions?

From:

LR <lruss@superlink.net>

Newsgroups:

comp.lang.c++

Date:

Tue, 08 Jan 2008 12:17:35 -0500

Message-ID:

<4783af9c$0$16302$cc2e38e6@news.uslec.net>

Kira Yamato wrote:

On 2008-01-08 09:18:53 -0500, LR <lruss@superlink.net> said:

Kira Yamato wrote:

Suppose class B is a subtype of class A, i.e.,
   class B : public A
   {
       ...
   };

But now how do I do the following kind of morphism? Suppose I have
functions
   void foo(const B &);
and
   void bar(const A &);
I like to be able to declare a function pointer
   void (*fp)(const B&);
and make "polymorphic" assignments like
   p = foo; // this is ok in C++.
   p = bar; // semantically this make sense, but C++ syntax won't
allow this!

Would it be acceptable to create a function,
void bar(const B &);

You mean
void foo(const B &);

No, that's not what I meant. I don't think that I made myself clear, so
I'll try again.

Untested, and not complete...

This is what I think you started with.
class A {
};
class B : public A {
};
void foo(const B &) {
}
void bar(const A &) {
}

now I suggest adding this function
void bar(const B &b) {
     ::bar( reinterpret_cast<const A&>(b) );
}

so you can
     void (*fp)(const B&);
and
     B b;
     fp = bar; // will assign the address of void bar(const B &) to fp
     fp(b);
     fp = foo;
     fp(b);

Or else I didn't understand your question.

LR