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Home/ Questions/Q 6807261
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T19:48:37+00:00 2026-05-26T19:48:37+00:00

a simple variable test: #!/bin/bash N=0 ls -l | while read L; do N=$((N+1))

  • 0

a simple variable test:

#!/bin/bash

N=0

ls -l | while read L; do
 N=$((N+1))
 echo $N
done

echo "total $N"

ran it then output:

1
2
3

total 0

i expected final N=3: “total 3”, but why did the value reset to 0 after the loop?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T19:48:37+00:00Added an answer on May 26, 2026 at 7:48 pm

    bash runs each statement in a pipe in its own subshell. (For external commands such as ls the subshell simply execs the command.) This effectively makes all of the variables local. You generally have to work around this by using redirection or command substitution instead of a pipe.

    EDIT: This seems to work:

    #!/bin/bash
    IFS=
    N=0
    
    while read L; do
      N=$((N+1))
      echo $N
    done <<<$(ls -l)
    
    echo "total $N"
    

    EDIT: As of bash 4.2 you can use shopt -s lastpipe in your script to disable the subshell for the last command in a pipe, which would then allow the rest of the original code to work as desired.

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