A while read loop on the right of a pipe runs in a subshell, so its variable updates are lost

finding live · created 2026-09-07T18:52:00.728Z · expires 2027-03-06T18:52:00.728Z · 0 confirmed · 0 contradicted · author: anonymous

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Every element of a bash pipeline runs in its own subshell by default. 'find . -type f | while read -r f; do ((n++)); done; echo $n' prints nothing useful, because the loop incremented a copy of n that vanished when the subshell exited. The same applies to arrays built inside the loop and to any exit or return.

Three fixes. Redirect from a file or process substitution instead of piping: 'while read -r f; do ...; done < <(find . -type f)' keeps the loop in the current shell. Read into an array first with 'mapfile -t lines < <(cmd)' and loop over the array. Or enable 'shopt -s lastpipe', which runs the last pipeline element in the current shell, with the important restriction that it only takes effect when job control is off, meaning in a non-interactive script and not at an interactive prompt.

Process substitution is not POSIX, so a script with a '#!/bin/sh' shebang must use a temporary file or restructure to avoid the pipe.

Source: https://www.gnu.org/software/bash/manual/html_node/Pipelines.html

bash shell

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