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I have a problem in one of my shell scripts. Asked a few colleagues, but they all just shake their heads (after some scratching), so I've come here for an answer.

According to my understanding the following shell script should print "Count is 5" as the last line. Except it doesn't. It prints "Count is 0". If the "while read" is replaced with any other kind of loop, it works just fine. Here's the script:

echo "1">
echo "2">>
echo "3">>
echo "4">>
echo "5">>


cat | while read ;
  let CNT++;
  echo "Counting to $CNT"
echo "Count is $CNT"

Why does this happen and how can I prevent it? I've tried this in Debian Lenny and Squeeze, same result (i.e. bash 3.2.39 and bash 4.1.5. I fully admit to not being a shell script wizard, so any pointers would be appreciated.

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up vote 20 down vote accepted

BASh FAQ entry #24: "I set variables in a loop. Why do they suddenly disappear after the loop terminates? Or, why can't I pipe data to read?"

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You get the bonus, since your answer provided me with the widest range of options. – wolfgangsz Apr 13 '11 at 17:22
awesome! thanks! – arod Jul 6 '15 at 22:52

This is kind of a 'common' mistake. Pipes create SubShells, so the while read is running on a different shell than your script, that makes your CNT variable never changes (only the one inside the pipe subshell).

Group the last echo with the subshell while to fix it (there are many other way to fix it, this is one. Iain and Ignacio's answers have others.)


 cat | ( while read 
  let CNT++;
  echo "Counting to $CNT"
echo "Count is $CNT" )

Long explanation:

  1. You declare CNT on your script to be value 0;
  2. A SubShell is started on the | to while read;
  3. Your $CNT variable is exported to the SubShell with value 0;
  4. The SubShell counts and increase the CNT value to 5;
  5. SubShell ends, variables and values are destroyed (they don't get back to the calling process/script).
  6. You echo your original CNT value of 0.
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First shell script I ever wrote gave me the same issues, banged my head against the wall for awhile before finding out that that pipes spawn additional shells. Any variable you mess with in a pipe will go out of scope as soon as the pipe ends--meaning that if you really, really want to do something with a variable outside of the pipe in which it was used, you'll have to hold state through something funky like a temporary file. – photoionized Apr 13 '11 at 17:17
Excellent answer, unfortunately I can only give one acceptance bonus. Sorry. – wolfgangsz Apr 13 '11 at 17:23

This works


while read ;
  let CNT++;
  echo "Counting to $CNT"
done <
echo "Count is $CNT"
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I like that, the smart way because you know where the needed data is and only need to get it back. If you dont know high skill solutions, you can always "read a file" hahahha. +1 for you. – erm3nda Jan 21 '15 at 8:11
Anyone reading this, be aware that the solution provided by Iain only works when you have your script explicitly invoking bash, by having the first line: #!/bin/bash and that: #!/bin/sh will not work. – Roadowl May 9 '15 at 0:08
Interesting, first example I ever saw where a Useless Use of Cat actually prevented code from working. By the way @Roadowl, the only bashism here is the line let CNT++ which should instead be CNT="$((CNT+1))" to make use of POSIX-compliant arithmetic expansion. The rest of it is portable already. – Wildcard Jan 12 at 16:31

Try passing the data in a sub-shell instead, like it's a file before the while loop. This is similar to lain's solution, but assumes you don't want some intermittent file:

while read var
  echo "variable: $var"
done < <(echo 45) #output from a command, script, or function
echo "total: $total"
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