[Courses] [C] Debugging (Re: help me find my C bug? (long))

[Conor Daly]

Arrgh!

I was just looking over these posts and realised that I have given the
"sign=" lines wrongly.  The correct version is:

sign = sqrt (amount * amount) / amount;

This will produce +1 for positive numbers and -1 for negative.

On Wed, Jul 10, 2002 at 12:56:03PM +0000 or so it is rumoured hereabouts, 
Conor Daly thought:
> 
> On the matter of floating point imprecision, we run a database of weather
> records here and record quite a bit of stuff to one place of decimals.
> However, at times, a search for "rain=3.3" returns no data while "rain >3.29
> and rain <3.31" will return hundreds of lines!
> 
> Using 
> 
> total=(amount * 100) + 0.5;
> 
> works, but to cope with negative numbers you need just a little more.  You
> can use two (probably more) different approaches;
> 
> 1. if else
> 
> if ( amount < 0 ) {
>   total=(int) (amount * 100) - 0.5);
> } else {
>   total=(int) (amount * 100) + 0.5);
> }
> 
> 2. use the sign
> 
> sign = sqrt (amount * amount);

sign = sqrt (amount * amount) / amount;

> total=(int) (amount * 100) + (sign * 0.5);
> 
> note that either of these methods give rise to an inconsistency where -0.5
> rounds to 0 while -1.5 rounds to -2 and -1 gets missed along the way.  The
> solution to this (IMO!) is to use just a *little* extra for the rounding
> figure.  ie.
> 
> 1. if else
> 
> if ( amount < 0 ) {
>   total=(int) (amount * 100) - 0.50001);
> } else {
>   total=(int) (amount * 100) + 0.5);
> }
> 
> 2. use the sign
> 
> sign = sqrt (amount * amount);

sign = sqrt (amount * amount) / amount;

> total=(int) (amount * 100) + (sign * 0.50001);
> 
> This will produce expected behaviour but ONLY so long as the "little bit
> extra" is well below the precision of the numbers you are rounding.
> 
> Conor (rambling...)

Conor
-- 
Conor Daly <conor.daly at oceanfree.net>

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