Re: bash and real arithmetic

[Igor Pechtchanski <pechtcha at cs dot nyu dot edu>]

On Fri, 11 Apr 2003, Randall R Schulz wrote:

> At 10:21 2003-04-11, Igor Pechtchanski wrote:
> >On Sat, 12 Apr 2003, Stepan Yakovenko wrote:
> >
> > > HI!
> > >
> > >   Is it possible to get for loops working
> > >   in double arithmetic in bash ? I've got to run
> > >   a program with a real argument many times, changing it
> > >   every run by some fixed small (<1) number.
> > >
> > >   Thanx in advance, sorry for my broken English.
> >
> >Umm, scale?  For example:
> >
> >    for (( i=0; i < 10; i++ )); do echo 0.$i; done
>
> Umm, leading zero suppression?
>
> % for (( i=0; i < 100; i++ )); do echo 0.$i; done
>
> 0.0
> 0.1
> [snip]
> 0.20
> etc.
> Randall

Yes, yes...  It was an /example/.  If you want leading zeros for numbers
greater than 10, use bash's "printf" builtin:

   for (( i=0; i < 100; i++ )); do echo 0.$(printf "%02d" $i); done

Better? :-p
	Igor
P.S. You're right, though, I should've warned him about it...
-- 
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