From:	Hans-Bernhard Broeker <broeker AT physik DOT rwth-aachen DOT de>
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: How would I build a pointer to a method of my class
Date:	10 Apr 2001 11:09:41 GMT
Organization:	Aachen University of Technology (RWTH)
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Wormy <wormy AT technologist DOT com> wrote:
> -------------------------------------------
> typedef void (My_func)(short parameter);

> My_func Pointer_to_My_func;
> -------------------------------------------

> and then, later in my code (DOTHIS_func and DOTHAT_func are declared and
> defined somewhere)

> -------------------------------------------
> if (i == 0)
>     Pointer_to_My_func = DOTHIS_func;
> else
>     Pointer_to_My_func = DOTHAT_func;

> (*Pointer_to_My_func)(parameter_to_pass);
> ----------------------------------------------

> OK.... but now, how am I doing it in C++ if I'm fumbling around with a
> class?

The "typical" OO-style method would be to not use a function pointer,
but rather a special-purpose class. I.e. you'ld set up a purely
abstract base class (don't sue me if the syntax is a bit off...):

	class my_handlertye {
	   void handlermethod(short parameter) = 0;
	}

Every handler function would be turned into a class derived from
my_handlertype, and then implement handlermethod(). Code using the
function pointer would instead use a pointer to an object of type
my_handlertype, and call handlermethod where needed:

	my_handlertype * handler;

	if (condition)
	   handler = new handler_type_1;
	else
	   handler = new handler_type_0;

	/* ... */


	handler->handlermethod(argument);

-- 
Hans-Bernhard Broeker (broeker AT physik DOT rwth-aachen DOT de)
Even if all the snow were burnt, ashes would remain.

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