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ShellCheck plugin for SonarQube

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SC2106: 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.