Path: chuka.playstation.co.uk!scea!greg_labrec@interactive.sony.com
From: Joshua Meeds <dreamer@sinclair.net>
Newsgroups: scee.yaroze.programming.gnu_compiler
Subject: Re: printf works, code doesn't
Date: Wed, 15 Apr 1998 19:42:28 -0700
Organization: SCEA News Server
Lines: 51
Message-ID: <35357014.E195CEDC@sinclair.net>
References: <3532D4BA.49B30CA9@sinclair.net> <3534730B.2417@127.0.0.1>
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Well, that was the actual code, but that line wasn't remmed out (I think
it was my newsgroup poster that messed it up).  You know what fixed the
problem?  I had to type cast the equation as a double, like so:

Fun=(double)(((Y2-Y1) / (X2-X1));

Majik wrote:
> 
> > double Slope(int a,int b)
> > {
> >   u_long X1,Y1,X2,Y2;
> >   double Fun;
> >
> >   X1=Dude[a].position.vx;
> >   Y1=Dude[a].position.vz;
> >   X2=Dude[b].position.vx;
> >   Y2=Dude[b].position.vz;
> >   printf2("\n%d / %d = %e -- ",Y2-Y1,X2-X1, (Y2-Y1) / (X2-X1));
> > //printf2 = printf that supports %e   Fun=((Y2-Y1) / (X2-X1));
> >   printf("%e",Fun);
> >
> >   return (Fun);  // Used to be return ((Y2-Y1) / (X2-X1));  - Same thing happened
> > }
> 
> Is this the actaul code you are using? I only ask because 'Fun' isnt
> assigned any
> value, it is hidden by the // comment, what happens if you do
> 
> //printf2 = printf that supports %e
> Fun=((Y2-Y1) / (X2-X1));
> 
> instead of
> 
> > //printf2 = printf that supports %e   Fun=((Y2-Y1) / (X2-X1));
> 
> or is it just my mailreader that has screwed that bit up, sorry if I've
> been patronising.
> 
> M.
> 
> --
> -- Majik - Owner/Curator of the Sinclair Microcomputer Museum
> --      Spod in a suitcase - ' Have Compiler, will travel '
> --              http://mudhole.spodnet.uk.com/~majik/
> --

-- 
 
                - Dreamwriter Dragon
                  dreamer@sinclair.net
                        -==UDIC==-