Re: declare in a function makes a variable unable to be found with declare -p in some cases

[Chet Ramey]

On 2/16/15 3:38 PM, SN wrote:
> Hello all,
> 
> I have found a problem with the declare builtin.
> 
> Patch Level: 33
> Release Status: release
> 
> Description:
>         Apparently, there is a problem with how bash interprets some
> variable assignments.
>         It only happens in a function (probably related to `declare'
> making variables local).

Yes, that's part of it.

> Repeat-By:
>         # OK
>         $ x() { declare -a var=(); declare -p var; }; x
>         declare -a var='()'
>         # not OK
>         $ y() { declare -a var='()'; declare -p var; }; y
>         bash: declare: var: not found
> 
>         Note that the format used in y is what `declare -p' displays.

Yes. The issue is that the assignment doesn't make the variable visible,
when it clearly should.  I've attached a patch that will fix that.

The next version of bash will do things differently, as noted in a couple
of followups to this message.  The proposal that prompted the changes is
part of a long thread from December:

http://lists.gnu.org/archive/html/bug-bash/2014-12/msg00115.html

Chet
-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer
                 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU    chet@case.edu    http://cnswww.cns.cwru.edu/~chet/

compound-array-assign-visible.patch
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