Sender:	nate AT cartsys DOT com
Message-ID:	<358EBF59.DAC4952C@cartsys.com>
Date:	Mon, 22 Jun 1998 13:32:25 -0700
From:	Nate Eldredge <nate AT cartsys DOT com>
MIME-Version:	1.0
To:	djgpp AT delorie DOT com
Subject:	Re: Casting void pointers
References:	<6mkaos$k7o AT dfw-ixnews6 DOT ix DOT netcom DOT com> <6mkfnv$hcr AT espresso DOT cafe DOT net> <358DECE1 DOT 67C137A4 AT alcyone DOT com>

Erik Max Francis wrote:
> 
> Kaz Kylheku wrote:
> 
> > You can create a typedef name for the function pointer and then use
> > the parenthesized typedef name. Or you can write cast expressions
> > like:
> >
> >         void *q = 0;
> >         double (*p)(double, int) = (double (*)(double, int)) q;
> 
> I think the original poster was really looking for the answer to the
> question:  For a given declaration, how do I determine its type (for
> casting)?
> 
> The answer is that if you remove the identifier name from the
> declaration, that's the type.  So in the declaration int p, int is the
> type; in the declaration char *s, char * is the type, and in the
> declaration double (*p)(double, int), double (*)(double, int) is the
> type, even though it looks a little strange.

#ifdef PLUG
I'd like to mention that I have ported a program called `cdecl', whose
job it is to covert casts, declarations, etc., to and from readable
English.  You can find it at my web site:

http://www.cartsys.com/eldredge/
#endif
-- 

Nate Eldredge
nate AT cartsys DOT com

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