On Tue, Aug 23, 2016 at 2:34 AM, Weshakie Löwe <wesha...@gmail.com> wrote:
> When storing the value of code executed in a subshell the return value is
> always 0 if the variable is local.
>
> Code example:
>
> A(){
> local return_value="$(bash -c "exit 1")"
> echo $?
> }
>
> function A: returns 0 - even though obviously the return value is 1.
>
``local'' itself is a command. So ``local var=value'' would usually return
0 because it _successfully_ assigns the value the the var. ``local'' would
return non-zero when there's something wrong with the syntax. For example
``local invalid.var.name'' or ``readonly var=1; local var=2''.
>
> B(){
> return_value="$(bash -c "exit 1")"
> echo $?
> }
>
> function B: returns 1 - as expected, the only difference to function A
> being not using a local variable to store the value.
>
Without ``local'', the value of $? would be changed by the $(...) part in
the assignment.
-clark
>
>
>
> Best regards,
>
> A happy bash user
>
>
>
