Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!att!dptg!ulysses!andante!alice!ark
From: ark@alice.UUCP (Andrew Koenig)
Newsgroups: comp.lang.c++
Subject: Re: pointers to member functions
Message-ID: <10206@alice.UUCP>
Date: 2 Dec 89 15:31:46 GMT
References: <950@sl10c.concurrent.co.uk>
Organization: AT&T Bell Laboratories, Liberty Corner NJ
Lines: 46

In article <950@sl10c.concurrent.co.uk>, asc@concurrent.co.uk (Andy Chittenden) writes:

[example that declares `dispatcher' as taking a pointer to a member of
class `base'

> 	dispatcher a(base::fn1);
> 	dispatcher b(derived::fn2);	// fails to compile under v2

> // I can understand why the second line does not compile (derived::fn2
> // is not a function of base).  However, how is one meant to achieve the
> // above effect under v2 (pass an arbitrary function of a derived class
> // with matching parameters)?

The `dispatcher' function takes a pointer to a member of class `base.'
There may be some members of class `derived' that are not members of
class `base,' so you can't pass any member of `derived' to a function
that takes a pointer to a member of `base' as its argument.

In particular, if fn2 were a member of `base' you would have been able
to say

	dispatcher b(base::fn2);

and there would have been no problem.

Watch this, though.  Every member of class `base' is also a member of
class `derived.'  Therefore you CAN pass a member of class `base'
to a function that expects a pointer to a member of class `derived'!
Example:

	void f(int (derived::*)(int));

	main()
	{
		f(derived::fn2);	// OK
		f(base::fn1);		// also OK!
	}

This may seem backwards at first, but if you think about it for a while
you will see that it has to be this way, version 1 to the contrary
notwithstanding.

This phenomenon is often called `contravariance.'
-- 
				--Andrew Koenig
				  ark@europa.att.com