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This Bash snippet works as expected:
$ fun1() { x=$(false); echo "exit code: $?"; } $ fun1 exit code: 1 But this one, using local, does not as I would have expected:
$ fun2() { local x=$(false); echo "exit code: $?"; } $ fun2 exit code: 0 Can anyone explain why does local sweep the return code of the command?
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Extracting the elusive exit code To display the exit code for the last command you ran on the command line, use the following command: $ echo $?
How to get the exit code of a command. To get the exit code of a command type echo $? at the command prompt. In the following example a file is printed to the terminal using the cat command.
Overview. In Linux, when a process is terminated, it returns an exit code. Upon successful execution, this code is equal to zero. Any non-zero exit code indicates that some error occurred.
false returns an exit status value of 1 (failure). It ignores any arguments given on the command line. This option can be useful in shell scripts.
The reason the code with local returns 0 is because $? "Expands to the exit status of the most recently executed foreground pipeline." Thus $? is returning the success of local
You can fix this behavior by separating the declaration of x from the initialization of x like so:
$ fun() { local x; x=$(false); echo "exit code: $?"; }; fun exit code: 1 If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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