Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!wuarchive!udel!haven!uvaarpa!murdoch!astsun9.astro.Virginia.EDU!cak3g
From: cak3g@astsun9.astro.Virginia.EDU (Colin Klipsch)
Newsgroups: comp.sys.mac.programmer
Subject: Re: Why can't the Mac add?
Summary: Don't use FP indexing
Message-ID: <1990Sep24.160108.26148@murdoch.acc.Virginia.EDU>
Date: 24 Sep 90 16:01:08 GMT
References: <45060@apple.Apple.COM>
Sender: news@murdoch.acc.Virginia.EDU
Organization: University of Virginia
Lines: 64

In article <45060@apple.Apple.COM> das@Apple.COM (David Shayer) writes:
>I was running this simple program.
>main ()
>{
>	float	x; 
>	for (x=0.0;x!=10.0;x+=0.2)
>		printf ("x=%f \n",x);
>}

>As you can see,  it ought to stop when x==10.0.  However, 
>it actually runs in an infinite loop.  This is because
>x never equals exactly 10.0.

>I have a Mac IIci, so I have a 68882 FPU.  This program
>fails in both Think and MPW C.  I didn't compile with any
>special FPU options on.  A friend ran it on a
>PC clone, and it failed in the same way there.

>Please enlighten me.

Never, ever, EVER use a floating point variable as a loop index!  Not in
any language, on any computer, under any circumstances.  Do not use
them in a box, do not use them with a fox.  (This, of course, is my
vehement opinion.)

Always use integers for the very reason you've discovered: floating
point numbers suffer from round-off error for most fractions, and that
certainly includes powers of ten.  0.2 represented in binary is
infinitely repeating.  The computer can't carry around an infinite
sequence of bits, so it's loop step is not REALLY 0.2, but the closest
binary approximation.  (The "double" approximation will be closer
than the "float", but will still be unequal.)

This is not a defect of your Mac, or the PC.  It's a result of the
fact that computers work in binary, not decimal.  You can usually
get away with using floating point numbers in loops IF the variable
AND the step size are exactly integers, but it's still a questionable
policy.

To work around this problem, you could use:

   for (x = 0.0; x <= 10.0; x += 0.2) {}

But the best solution:

   int i;
   for (i = 0; i <= 50; i++) { x = i/5.0; ... }

This more faithfully generates the results you (presumably) wanted:
to have x go from 0 to 10 in 50 evenly distributed intervals.  More
importantly, your loop is guaranteed to terminate, and it will
terminate after the correct number of steps.

Hope this helps.

--------------------------------------------------------------------------
"May the forces of evil become confused on the way to your house."
                                                          -- George Carlin
Bemusedly,                   | Disclaimers:
  Colin Klipsch              |   Not guaranteed to fulfill any purpose,
  Property of UVa Ast. Dept. |   express or implied.  Contents may have
  Charlottesville, Virginia  |   settled during shipping.  Not rated.  May
  cak3g@virginia.edu         |   cause drowsiness.  Use before 29-Feb.
____________________________/ \___________________________________________