This Bash snippet works as I would\'ve expected:
$ fun1() { x=$(false); echo \"exit code: $?\"; }
$ fun1
exit code: 1
But this one, using local
, does not:
$ fun2() { local x=$(false); echo \"exit code: $?\"; }
$ fun2
exit code: 0
Can anyone explain why does local
sweep the return code of the command?
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
The return code of the local
command obscures the return code of false