1

This function should exit the calling script:

crash() {
  echo error
  exit 1
}

This works as expected:

echo before
crash
echo after         # execution never reaches here

But this does not:

echo before
x=$(crash)         # nothing is printed, and execution continues
echo after         # this is printed

How do I capture the result of a function, as well as allow it to exit?

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2
  • 1
    Find info about trap. Commented Mar 22, 2020 at 10:12
  • @LinuxSecurityFreak Thanks good idea Commented Mar 22, 2020 at 11:47

2 Answers 2

Reset to default
5

This is because $(crash) executes crash in a subshell, so the exit applies to the subshell and not to your script.

What is the point of capturing the output in a variable if you won't use it because the script exited anyway?

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1
  • 1
    Sometimes I want to return something, OR if something bad happens, to exit. Thanks for the explanation. Commented Mar 22, 2020 at 11:46
1

This should solve your problem:

echo before
x=$(crash) || exit       # if crash give -gt 0 value then exit with the same value
echo after
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2
  • 1
    exit (without $?) will exit with the value of the last command executed Commented Mar 22, 2020 at 13:16
  • While this may solve the OP’s problem, it doesn’t, strictly speaking, answer the question. This will exit if the function does return 1 and not if it does exit 0. Commented Mar 23, 2020 at 20:45

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