From:	"Steve Dondley" <stevedondley AT mediaone DOT net>
Newsgroups:	comp.os.msdos.djgpp
References:	<wyZ97.2407$n_3 DOT 3163465 AT typhoon DOT ne DOT mediaone DOT net> <9k9u34$d9c$1 AT slb7 DOT atl DOT mindspring DOT net>
Subject:	Re: Type casting and pointers
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Date:	Wed, 01 Aug 2001 22:13:41 GMT
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To:	djgpp AT delorie DOT com
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Reply-To:	djgpp AT delorie DOT com

Marp,

Great, thanks so much.  One othe question, though.  The book doesn't cover
the "float *" and "float **" casts.  The book just weirdly throws it into
this example with no explanation.  Must be an editor's mistake. Where can I
learn about this?  I've tried looking this up on various web C tutorials but
haven't been able to find it.


"Marp" <marp AT 0 DOT 0 DOT 0 DOT 0> wrote in message
news:9k9u34$d9c$1 AT slb7 DOT atl DOT mindspring DOT net...
> The only purpose of typecasting in the example given is to suppress a
> compiler warning. It's not a good example :-) Normally you would use
> typecasting to override the compiler's default behavior in promoting data
> types.
>
> Given this:
>
> int a = 1;
> double b = 2.5;
> double c;
>
> And this:
>
> c = a + b;
>
> The variable c will hold 3.5 because the compiler promotes a to double for
> the purposes of the addition.
>
> But this:
>
> c = a + (int)b;
>
> The variable c will hold 3 since you told it to demote b to int (which
> causes it to use the truncated value of b for addition).
>
> BTW, you get a warning in the book's example without a cast because you
are
> assigning a float ** to a float *. They are not the same thing.
>
> Hope this helps.
>
> - Marp
>
> "Steve Dondley" <stevedondley AT mediaone DOT net> wrote in message
> news:wyZ97.2407$n_3 DOT 3163465 AT typhoon DOT ne DOT mediaone DOT net...
> > Hello,
> >
> > I've got a newbie C question.  I've got a beginner's C book that doesn't
> > explain something very well. Example code from the book:
> >
> > float *p;
> > float balance[10][5];
> > ...assignment stuff here...
> > p = (float *) balance;
> >
> > My code works just fine without the (float *) type cast, but I do get a
> > warning.  Why exactly do you need it?  Doesn't declaring *p as a float
> > pointer at the top of the source take care of that?  The book also never
> > explains what exactly the asterisk in (float *) is.  I'm just assuming
it
> > type casts the pointer.
> >
> > Can someone please provide some elucidation?  Thanks!
> >
> > ---Steve
> >
> >
> >
>
>

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