shell variables becoming zero outside the loop

cool244 · 05-18-2006, 07:14 PM

Folks,

I have a simple script, the calcualtions done inside a while loop or lost outside the loop. Here is the script.

Code:

#! /bin/ksh
vmstat 1 6 | sed '1,4d' | while read a b c d e f g h i j k l m n o p junk
do
  echo "CNT is $cnt"
  if [[ $cnt -le 3 ]]; then
    echo "in loop"
    let sum1=$sum1+$a
    let sum2=$sum2+$b
    let sum3=$sum3+$c
    let sum4=$sum4+$d
    echo "###inside if loop values are sum1=${sum1} sum2=${sum2} sume3=${sum3} sum4=${sum4}###"
  fi
    let cnt=$cnt+1
    echo "###outside if loop values are sum1=${sum1} sum2=${sum2} sume3=${sum3} sum4=${sum4}###"
done
    echo "###outside while loop values are sum1=${sum1} sum2=${sum2} sume3=${sum3} sum4=${sum4}###"

The output:/tmp/test1

Code:

CNT is
in loop
###inside if loop values are sum1=0 sum2=0 sume3=0 sum4=10129064###
###outside if loop values are sum1=0 sum2=0 sume3=0 sum4=10129064###
CNT is 1
in loop
###inside if loop values are sum1=0 sum2=0 sume3=0 sum4=20258128###
###outside if loop values are sum1=0 sum2=0 sume3=0 sum4=20258128###
CNT is 2
in loop
###inside if loop values are sum1=0 sum2=0 sume3=0 sum4=30387160###
###outside if loop values are sum1=0 sum2=0 sume3=0 sum4=30387160###
CNT is 3
in loop
###inside if loop values are sum1=0 sum2=1 sume3=0 sum4=40516192###
###outside if loop values are sum1=0 sum2=1 sume3=0 sum4=40516192###
###outside while loop values are sum1= sum2= sume3= sum4=###

As you can sum4 is NULL outside the loop. The same script works in AIX. Can someone please let me know, what is wrong here? I tried both ksh and bash and results are same.

Thank you

burninGpi · 05-18-2006, 09:48 PM

I'm pretty sure this belongs in PROGRAMMING...

XavierP · 05-20-2006, 07:14 AM

Moved: This thread is more suitable in Programming and has been moved accordingly to help your thread/question get the exposure it deserves.

randyding · 05-20-2006, 01:09 PM

Its important to realize that piping '|' causes the shell to startup another instance of the shell (ksh in your case) and pipes the data to its stdin where "read a..junk" picks it up.
When this "subshell" exits, all of its environment is destroyed. So, variables you set outside are readable only inside the subshell, and when the shell exits any modifications you made to the read-only variables are lost. This behavior is normal.

One way to get around this is to write the output generated by the subshell to a temporary file and after the code block read the temp file back in. Another way is to put the whole code block inside a output=`<codeblock>` and echo the results.

I'm interested in hearing what other programmers do, because this is frequently a frustration for myself as well.

spirit receiver · 05-20-2006, 03:33 PM

Quote:

Originally Posted by randyding

Its important to realize that piping '|' causes the shell to startup another instance of the shell (ksh in your case) and pipes the data to its stdin where "read a..junk" picks it up.

So you'll either have to return the values from the subshell as randyding proposes (although there are probably more beautiful solutions than using a temporary file), or you simply do all the output from the subshell as in the following example (note the parentheses):

Code:

#! /bin/ksh

vmstat 1 6 | sed '1,4d' | 
(
    while read a b c d e f g h i j k l m n o p junk
      do
      echo "CNT is $cnt"
      if [[ $cnt -le 3 ]]; then
	  echo "in loop"
	  let sum1=$sum1+$a
	  let sum2=$sum2+$b
	  let sum3=$sum3+$c
	  let sum4=$sum4+$d
	  echo "###inside if loop values are sum1=${sum1} sum2=${sum2} sume3=${sum3} sum4=${sum4}###"
      fi
      let cnt=$cnt+1
      echo "###outside if loop values are sum1=${sum1} sum2=${sum2} sume3=${sum3} sum4=${sum4}###"
    done
    echo "###outside while loop values are sum1=${sum1} sum2=${sum2} sume3=${sum3} sum4=${sum4}###"
)