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

[Conor Daly]

On Wed, Jul 10, 2002 at 05:36:24PM +1000 or thereabouts, Mary wrote:
> On Tue, Jul 09, 2002 at 10:28:50PM -0700, Suzi Anvin wrote:
> > Hold on... pointers are chapters away yet!  :)  this is like my 3rd or
> > 4th day at this!  So far, all the programs I've been playing with are
> > one-function programs and I've yet to have explained to me (by the
> > book) the difference between global and local variables...  doing it
> > the book's way for now, sand I'm really hoping they eventually explain
> > WHY all the examples declare all the variables globally so far.  :)
> 
> That sounds... odd. Local variables in C aren't hard, you can declare
> them at the start of any block (blocks are bracketed by {} braces), and
> their scope is until the end of the block.

Heh, I got bitten by scope on one accasion recently.  I'd just installed
vim6.0 to take advantage of "folding" (It's brilliant, 80 line functions
hidden behind one line!).  I wanted to make loops, if else ladders, variable
declarations all disappear and, given that {} is the usual delimeter for the
folding function in vim, I placed all of my variable declarations inside {}
pairs.  Then I wondered why my code wouldn't compile.  My variables had all
become limited in their scope to just the {} blocks where they were
_declared_!

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);
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);
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 Daly 
Met Eireann, Glasnevin Hill, Dublin 9, Ireland
Ph +353 1 8064276 Fax +353 1 8064247
------------------------------------
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