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From: minow@bolt.enet.dec.com (Martin Minow)
Newsgroups: comp.sys.mac.programmer
Subject: Re: C++ for a beginner
Message-ID: <20434@shlump.nac.dec.com>
Date: 21 Feb 91 22:12:05 GMT
Sender: newsdaemon@shlump.nac.dec.com
Distribution: comp
Organization: Digital Equipment Corporation
Lines: 39


In article <1991Feb21.012535.2899@vax5.cit.cornell.edu>,
a5uy@vax5.cit.cornell.edu writes...

>Here's a first obviously basic question...but a good answer would clear up all
>sorts of fog... All the sample progs have a main() that looks like this:
> 
>void   main(void)
>       gApplication = new(CPedestalApp);
>       ((CPedestalApp*)gApplication)->IPedestalApp();
>       gApplication->Run();
>       gApplication->Exit();
> 
> 
>My question is why does gApplication need to be type cast as a (CPedestalApp*)
>in order to send the message IPedestalApp() when, as far as I can tell, it's
>type *IS* CPedestalApp* ???  And assuming this type cast is necessary, why is
>it *NOT* necessary in order to send messages Run() and Exit() ???

Welcome to methods and prototyping.  If you define gApplication as
	extern CApplication *gApplication;
you need to cast it to a (CPedestalApp *) in order to send use a method
that is only defined for CPedestalApp's.  If, on the other hand, you're
sending a CApplication method, you don't need the cast.

What I've been doing is to define gApplication to my own application type:

	gApplication = (MyApplication *) new(MyApplication);
	gApplication->IMyApplication();
	gApplication->Run();
	gApplication->Exit();

You should always use "require prototypes" and "check prototypes"
on Think C, even if it is painful to cast things at times.

Hope this helps.

Martin Minow
minow@bolt.enet.dec.com