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  #1 (permalink)  
Old 02-07-2001
blaze blaze is offline
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Join Date: Feb 2001
Location: MN
Posts: 2
Question

Simple script trying to increment a counter within an echo statement never gets past 1 - PLEASE HELP!

Thanks.
~~~~~~~~~~~

#!/bin/sh
stepup()
{
STEP=`expr $STEP + 1`
echo $STEP
}

#
# Initialize variables
#
STEP=0

echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
  #2 (permalink)  
Old 02-07-2001
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PxT PxT is offline Forum Advisor  
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Join Date: Oct 2000
Location: Sacramento, CA
Posts: 909
the problem is that your function "stepup" gets evaluated at the beginning of execution. I would do something like this instead:


#!/bin/sh

stepup()
{
let STEP=$1+1
echo $STEP
}

STEP=0

STEP=`stepup $STEP`
echo "Counter: $STEP"
STEP=`stepup $STEP`
echo "Counter: $STEP"
STEP=`stepup $STEP`
echo "Counter: $STEP"
STEP=`stepup $STEP`
  #3 (permalink)  
Old 02-08-2001
blaze blaze is offline
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Join Date: Feb 2001
Location: MN
Posts: 2
Angry

Is there any way to evaluate the function "stepup" and echo/print it out in one statement or are we stuck doing 2 separate statements - one to evaluate the function and another to display it. Is there a way to do it within the function (ie, return)?

Thanks.
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