Variables in scripts

[Dave Basener]

Brian,
The script, exactly as you give it here fails with a syntax error in the 
test command because you initialize "change" to 0 but then use 
$changed.  This, in /bin/sh will always cause a syntax error because 
$changed has no value and the "-eq" is a binary operator.

I am unable to test your script on a Linux machine in true "sh" because 
all of my systems have /bin/sh linked to /bin/bash.  I was able to try 
it on a TRU-64 (osf5.1) UNIX machine with a real "bin/sh".

Results:
Original script fails in both bash and true sh

After taking care of the apparent typo by changing "change=0" to 
"changed=0" the script works in both environments.  The final echo emits 1.

If your typo was only in the mail and not in the real script, the I have 
no idea why it wouldn't work in /bin/sh on Linux. 

Dave Basener

Brian McGrew wrote:

>I'm asking this question here becuase what I have works on Solaris but not Redhat 7.3
>
>I have a simple script:
>
>#!/bin/sh
>
>change=0
>
>while [ $changed -eq 0 ];
>do
>	changed=1
>	echo $changed
>done
>
>echo $changed
>exit
>
>When I come to the last echo $changed, it's back to 0.  Any idea why that is?  This works find on Solaris???
>
>-brian
>
>Brian D. McGrew        { brian at doubledimension.com || brian at visionpro.com }
>---
>  
>
>>YOU!  Off my planet!
>>    
>>
>
>  
>


-- 
"... be the change you wish to see in the world."  - Gandhi

David Basener               http://www.aurora.edu/~dbasener
System Administrator                Dave.Basener at aurora.edu
Aurora University                              630 844 4889