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From: diamond@diamond.csl.sony.junet (Norman Diamond)
Newsgroups: comp.lang.c++
Subject: Re: Pointers to Methods Question
Message-ID: <10208@socslgw.csl.sony.JUNET>
Date: 2 May 89 09:40:00 GMT
References: <797@usl.usl.edu>
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Reply-To: diamond@csl.sony.junet (Norman Diamond)
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In article <797@usl.usl.edu> pcb@usl.usl.edu (Peter C. Bahrs) writes:

  >am not sure why I receive the results below? Any comments?

I >#include <stream.h>
N >
E >class What {
W >public:
S >  What (void) { cout << form("Construct What\n"); }
  >  void virtual print();
I >};
S >
  >void What::print()
A >{
N >  cout << form ("I am What's print routine\n");
  >}
A >
S >class Where : public What {
S >public:
H >  Where (void) { cout << form("Construct Where\n"); }
O >  void print(void) { cout << form ("Where's print routine\n"); }
L >};
E >
  >main ()
  >{
  >  typedef void What::WhatMember();
  >  typedef void Where::WhereMember();
  >  What* x = new What;
  >  Where* y = new Where; 
  >  
  >  WhatMember* fp1;
  >  WhereMember* fp2;
  >  fp1 = &Where::print;

Interesting.  I wonder why this isn't considered a type mismatch.

  >  (x->*fp1)();
  >  fp2 = &Where::print;
  >  (y->*fp2)();
  >}


  >Construct What
For your invocation of new What.
  >Construct What
  >Construct Where
Both for your invocation of new Where.
  >I am What's print routine
Since print() is virtual, (x->*fp1)() calls the version of print()
which is defined for the actual type of (*x), which is What.
  >Where's print routine
Similarly, the actual type of (*y) is Where.
If print() were non-virtual, then Where::print() would be called
both times.

Norman Diamond, Sony Computer Science Lab (diamond%csl.sony.jp@relay.cs.net)
  The above opinions are my own.   |  Why are programmers criticized for
  If they're also your opinions,   |  re-inventing the wheel, when car
  you're infringing my copyright.  |  manufacturers are praised for it?