2 SC2106

Joachim Ansorg edited this page 2021-11-12 19:33:23 +01:00

Table of Contents

• This only exits the subshell caused by the pipeline.
• Problematic code:
• Correct code:
• Rationale:
• Related resources:

This only exits the subshell caused by the pipeline.

Problematic code:

for i in a b c; do
  echo hi | grep -q bye | break
done

Correct code:

for i in a b c; do
  echo hi | grep -q bye || break
done

Rationale:

The most common cause of this issue is probably using a single | when || was intended. The reason this message appears, though, is that a construction like this, intended to surface a failure inside of a loop:

for i in a b c; do false | break; done; echo ${PIPESTATUS[@]}

may appear to work:

$ for i in a b c; do false | break; done; echo ${PIPESTATUS[@]}
1 0

What's actually happening, though, becomes clear if we add some echos; the entire loop completes, and the break has no effect.

$ for i in a b c; do echo $i; false | break; done; echo ${PIPESTATUS[@]}
a
b
c
1 0
$ for i in a b c; do false | break; echo $i; done; echo ${PIPESTATUS[@]}
a
b
c
0

Because bash processes pipelines by creating subshells, control statements like break only take effect in the subshell.

Related resources:

• Contrast with the related, but different, problem in this link.
• Bash Reference Manual: Pipelines, esp.:

  Each command in a pipeline is executed in its own subshell.

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