Re: Does object have function?

From:

"Daniel T." <daniel_t@earthlink.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 28 Oct 2010 21:02:04 -0400

Message-ID:

<daniel_t-BB2949.21020428102010@70-3-168-216.pools.spcsdns.net>

In article <iacrrl$ah6$1@four.albasani.net>,
"Jim Langston" <tazmaster@rocketmail.com> wrote:

What I am trying to accomplish: I have a map of polymorphic objects and
these objects may have certain methods defined or not. If the instance has
the method then I want to call it, otherwise not.

It is trivial to create a virtual function for a few functions. But I would
have to add every single function that could be interfaced.

I would just to somehow be able to tell if an instance has a function
available. Consider:

#include <iostream>

class Base {
public:
   virtual ~Base() {}
};

class DerivedOne : public Base {
public:
   void foo() { }
};

class DerivedTwo : public Base {
public:
   void foo() { }
};

class DerivedThree: public Base {
};

int main() {
   Base* bps[3];
   bps[0] = new DerivedOne();
   bps[1] = new DerivedTwo();
   bps[2] = new DerivedThree();

   for ( auto i = 0; i < 3; ++i ) {
      DerivedOne* thisOne = dynamic_cast<DerivedOne*>( bps[i] );
      if ( thisOne )
         thisOne->foo();
   }
}

I would have to dynamic_cast for every class that had a foo() defined to
execute all foos. Is there a way to do what I want?

This would work:

class Base {
public:
   virtual ~Base() {}
};

class Fooer {
public:
   virtual void foo() = 0;
};

class DerivedOne : public Base, public Fooer {
public:
   void foo() { cout << "DerivedOne::foo()\n"; }
};

class DerivedTwo : public Base, public Fooer {
public:
   void foo() { cout << "DerivedTwo::foo()\n"; }
};

class DerivedThree: public Base {
};

int main() {
   Base* bps[3];
   bps[0] = new DerivedOne();
   bps[1] = new DerivedTwo();
   bps[2] = new DerivedThree();

   for ( int i = 0; i < 3; ++i ) {
      Fooer* thisOne = dynamic_cast<Fooer*>( bps[i] );
      if ( thisOne )
         thisOne->foo();
   }
}