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

[Piotr Grzybowski]

 after discussion with Grisha, the reason to different behaviour between:

f() { declare -n ref=var; declare -n ref; declare -p ref; }; f

and

f() { local var; declare -n ref=var; declare -n ref; declare -p ref; }; f

is:

in function context declare built-in always calls make_local_variable. this 
routine has no idea that the second declare re-declares already declared 
reference, and calls make_new_variable. the difference is: make_local_variable 
will not create a local variable when there is already one, hence everything 
seems fine. make_local_variable should check find_variable_noref(name) and act 
accordingly.
 It seems half-intended.

pg

On 30 Apr 2016, at 22:24, Grisha Levit wrote:

> I just re-built bash-20160415 snapshot and am observing the same behavior.  
> To clarify, the first case is the unexpected one -- shouldn't `declare -n 
> ref=var; declare -n ref' be a no-op, no matter if $var is set or not?  It is 
> a no-op when in global scope, but not inside a function.