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">input.data
echo "2">>input.data
echo "3">>input.data
echo "4">>input.data
echo "5">>input.data


cat input.data | 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.

6 Answers 6


See argument @ 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?" (most recently archived here).

Summary: This is only supported from bash 4.2 and up. You need to use different ways like command substitutions instead of a pipe if you are using bash.

  • You get the bonus, since your answer provided me with the widest range of options.
    – wolfgangsz
    Apr 13, 2011 at 17:22
  • 7
    The link is dead. This is why link-only answers are bad. At least summarise the answer here. Apr 29, 2017 at 19:38
  • god, yet another time where ksh is simply so much better... why, just why did everyone flock around bash. Dec 12, 2017 at 0:16
  • @FlorianHeigl: Are you claiming that ksh is the One True Shell? Dec 12, 2017 at 0:20
  • @IgnacioVazquez-Abrams no, but i claim that the while loop handling in bash is a horribly PITA. The loop handling was the long-runner keeping it from catching to 1993's functionality. The other things are getopt handling where the (also 1993) builtin handler was simple and capable, something you still can't get unless using i.e. docopt. I'm claiming that bash has put itself behind the curve for over 20 years, insistently and the amount of time spent on THIS THING HERE or millions of bad getopts uses is beyond measure - only accepted because most people will never know. Dec 12, 2017 at 0:28

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 input.data | ( 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.
  • 2
    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. Apr 13, 2011 at 17:17
  • Excellent answer, unfortunately I can only give one acceptance bonus. Sorry.
    – wolfgangsz
    Apr 13, 2011 at 17:23

This works


while read ;
  let CNT++;
  echo "Counting to $CNT"
done <input.data
echo "Count is $CNT"
  • I like that, the smart way because you know where the needed data is and only need to get it back. If you dont know high skill solutions, you can always "read a file" hahahha. +1 for you.
    – m3nda
    Jan 21, 2015 at 8:11
  • 1
    Anyone reading this, be aware that the solution provided by Iain only works when you have your script explicitly invoking bash, by having the first line: #!/bin/bash and that: #!/bin/sh will not work.
    – Roadowl
    May 9, 2015 at 0:08
  • 1
    Interesting, first example I ever saw where a Useless Use of Cat actually prevented code from working. By the way @Roadowl, the only bashism here is the line let CNT++ which should instead be CNT="$((CNT+1))" to make use of POSIX-compliant arithmetic expansion. The rest of it is portable already.
    – Wildcard
    Jan 12, 2016 at 16:31
  • 1
    What about when the loop is producing a string?
    – Ken Ingram
    Jan 24, 2020 at 3:22

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"

Another solution is just simply add shopt -s lastpipe before the while loop.

I say, if the trouble comes cause the while is in the last segment of the pipeline, and in Bash all the commands in a pipeline executes in a subshell in a separated process isolated one from the other, then, using the lastpipe will execute the last command in the pipeline in the foreground.

For example:

shopt -s lastpipe
cat input.data | while read ;

And almost everything stay the same.


I found a way using the stderr file to store the value of var i.

# reading lines of content from 2 files concatenated
# inside loop: write value of var i to stderr (before iteration)
# outside: read var i from stderr, has last iterative value

cat $f $g | \
while read -r s;
  echo $s > /dev/null;  # some work
  echo $i > 2
  let i++
read -r i < 2
echo $i
  • This is actually redirecting to a file named "2", not to the stderr.
    – hugomg
    Aug 6, 2020 at 23:33

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