Shell programming problems

[ettiennel]

Hi All

Attached is a script that sets two variables in two different loops. After the first loop the variables are set to their original values for some or other reason.

The second loop also sets the same variables using the same function but retains their values as set in the second loop. I am running SUSE 8 enterprise.

Does anyone perhaps know why the values are not set after the first loop?

#!/bin/bash

Change() {

   STRINGVAR="CHANGED"
   echo "$STRINGVAR + $INT"
   let INT=$INT+1
}

STRINGVAR="START"
typeset -i INT=0


ps -ef | while read line
do
   Change
done

echo "AFTER FIRST LOOP BLOCK = " $STRINGVAR
echo "AFTER FIRST LOOP BLOCK = " $INT

while [ $INT -le 10 ]
do
        Change
done

echo "AFTER SECOND LOOP BLOCK = " $STRINGVAR
echo "AFTER SECOND LOOP BLOCK = " $INT

OUTPUT
------------------------------------------------------------
CHANGED + 0
CHANGED + 1
CHANGED + 2
CHANGED + 3
.
.
.
CHANGED + 176
AFTER FIRST LOOP BLOCK = START
AFTER FIRST LOOP BLOCK = 0
CHANGED + 0
CHANGED + 1
CHANGED + 2
CHANGED + 3
CHANGED + 4
CHANGED + 5
CHANGED + 6
CHANGED + 7
CHANGED + 8
CHANGED + 9
CHANGED + 10
AFTER SECOND LOOP BLOCK = CHANGED
AFTER SECOND LOOP BLOCK = 11

[lakerdonald]

1. Please put code in [code] tags
2. I don't know why this is, I was under the impression that bash variables were global...
Oh well.

[sant]

Does anyone perhaps know why the values are not set after the first loop

Yes, that's because of the pipe. Every part of a pipe run inside a separate subshell. You can prove this trying:

ps -ef> $tm_file
while read line
do
Change
done< $tmp_file

Bye

[lakerdonald]

Yes, that's because of the pipe. Every part of a pipe run inside a separate subshell.

This isn't true on all shells on all Unices however, I don't think.