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SC2320

Joachim Ansorg edited this page 2022-10-31 15:27:34 +01:00

Table of Contents

• This $? refers to echo/printf, not a previous command. Assign to variable to avoid it being overwritten.
• Problematic code:
• Correct code:
• Rationale:
• Exceptions:
• Related resources:

This $? refers to echo/printf, not a previous command. Assign to variable to avoid it being overwritten.

Problematic code:

mycommand
echo "Command exited with $?"
if [ $? -ne 0 ]
then
  echo "Failed"
fi

Correct code:

mycommand
ret=$?
echo "Command exited with $ret"
if [ $ret -ne 0 ]
then
  echo "Failed"
fi

Rationale:

ShellCheck found a $? that always refers to echo or printf.

This most commonly happens when trying to show $? before doing something with it, without realizing that any such action will also overwrite $?.

In the problematic example, echo "Command exited with $?" was intended to show the exit code before acting on it, but the act of showing $? also overwrote it, so the condition is always false. The solution is to assign $? to a variable first, so that it can be used repeatedly.

Exceptions:

If you intentionally refer to echo to get the result of a write, you can ignore this message. Alternatively, write it out as in if echo $$ > "$pidfile"; then status=0; else status=1; fi

Related resources:

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