Bash: Variable type confusion

[Joel Hammer]

If I use a variable as an integer, then use as a string, it doesn't go
back to being an integer nicely. 
Oh well. Either I had forgotten about this behavior or I never ran across it
before. Almost the same thing at my age.
Joel


On Sun, Oct 20, 2002 at 05:59:16PM -0400, Joel Hammer wrote:
> What version of bash are you using?
> 
> I have having a lot of trouble with variable typing.
> My version of bash is:
> GNU bash, version 2.04.0(1)-release (i686-pc-linux-gnu)
> Copyright 1999 Free Software Foundation, Inc.   
> 
> 
> Joel
> 
> 
> On Sun, Oct 20, 2002 at 02:23:16PM -0700, Net Llama! wrote:
> > On 10/20/02 13:53, Joel Hammer wrote:
> > 
> > > I have always thought that bash does the right thing when it evaluates
> > > variables.
> > > Until today.
> > > This bit of script works as expected:
> > > k=0
> > > for i in `dir -1 *jpg`
> > > do
> > > k=$((k+1))
> > > echo $k
> > > done
> > >
> > > It prints out the numbers from 1 to whatever the number of jpg's in the
> > > directory.
> > >
> > > However, this fails:
> > > k=0
> > > for i in `dir -1 *jpg`
> > > do
> > > k=$((k+1))
> > > [ "$k" -eq 10 ] && k=x   <---The problem.
> > > echo $k
> > > done
> > 
> > Both of these work the same for me, using bash.
> > 
> > -- 
> > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> > L. Friedman                       	       netllama at linux-sxs.org
> > Linux Step-by-step & TyGeMo: 		    http://netllama.ipfox.com
> > 
> >    2:20pm  up 8 days,  2:38,  1 user,  load average: 0.11, 0.40, 0.41
> > 
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