Re: use local and $?

[Eric Blake]

On 01/12/2015 02:23 AM, l_j_f wrote:

>         local b=$(error); err=$?
>         echo b=$b err=$err
> }  
> main "$@" 
> 
> 3. the result
> -sh-4.3# ./test2.sh 
> a=ok err=0
> b=error err=0 #I think it should be "b=error err=1"

The code is behaving correctly, and only your expectations are wrong.
$() can only affect $? when it occurs in variable assignments in
isolation, but you used it as part of a 'local' command.  'local' is
documented as having exit status of 0 if it successfully assigned all
variables, and non-zero if it failed to set at least one variable.
Since b was successfully set, 'local' sets $? to 0, which overrides any
status in the embedded $().

You meant to write:

local b
b=$(error); err=$?

-- 
Eric Blake   eblake redhat com    +1-919-301-3266
Libvirt virtualization library http://libvirt.org

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