Re: declare [-+]n behavior on existing (chained) namerefs

[Chet Ramey]

On 4/28/16 9:49 PM, Grisha Levit wrote:

> In a slightly different version, with `declare -n r; r=a', the function
> exits with code 1 after the `r=a' statement:
> 
> $ declare -nt r=a; f() { declare a; declare -n r; r=a; declare -p a r; };
> f; echo $?
> 1

In this case, you create a self-referencing local nameref, which ends up
resolving to nothing, which causes an assignment error, which results in
function execution being aborted.  You create the self-referencing local
nameref because bash follows the nameref chain for `r', and resolves it
to `a'.  It's as Piotr surmised: as if you ran `typeset -n a; a=a'.

When asked to create a local variable that shadows a nameref from a
different context, declare needs to throw away the results of that nameref
chain search and just create a local variable.

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/