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BASH variable scope inside while loop and SHLVL (Re: FAQ E4)


From: Kevin R. Bulgrien
Subject: BASH variable scope inside while loop and SHLVL (Re: FAQ E4)
Date: Tue, 29 May 2001 14:12:07 -0500

From: http://mail.gnu.org/pipermail/bug-bash/2000-August/001026.html

> Scope of variable in shell script 
> 
> Chet Ramey address@hidden 
> Thu, 31 Aug 2000 15:31:17 -0400 
> 
> Previous message: Scope of variable in shell script 
> Next message: Bug in getopts builtin. 
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>
>> uname output: Linux pckvc 2.2.14-5.0 #1 Tue Mar 7 21:07:39 EST 2000 i686
>> 
>> Bash Version: 1.14
>> Patch Level: 7
>> 
>> Description:
>> 
>> Scope of variable "counter" in following constructions seems wrong.
> 
> The while loop is run in a subshell because of the pipeline. A subshell
> cannot affect its parent's environemnt. Read the Bash FAQ, question E4,
> for more details.

I have read the FAQ, and understand the concept, however, consider the
following shell transcript:

  $ bash --version
  GNU bash, version 2.04.11(1)-release (i386-redhat-linux-gnu)
  Copyright 1999 Free Software Foundation, Inc.
  $cat myscript
  echo $SHLVL
  echo "-"
  echo | while read var
  do
    echo $SHLVL
  done
  echo "-"
  echo $SHLVL
  $bash myscript
  2
  -
  2
  -
  2
  $

Given your explanation and reference to FAQ E4, should this script
not return:

  2
  -
  3
  -
  2

> -- 
> ``The lyf so short, the craft so long to lerne.'' - Chaucer
> ( ``Discere est Dolere'' -- chet)
> 
> Chet Ramey, CWRU address@hidden http://cnswww.cns.cwru.edu/~chet/
>

-- 

Kevin R. Bulgrien, Engineer
Mailto:address@hidden

Vertex RSI, A TriPoint Global Company       http://www.vcsd.com/
Controls & Structures Division 
1915 Harrison Road                          Tel: 903-295-1480 x288 
Longview, TX 75604-5438                     Fax: 903-295-1479





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