Re: "local -g" declaration references local var in enclosing scope

[Kerin Millar]

On Mon, 11 Mar 2024 11:45:17 -0400
Chet Ramey <chet.ramey@case.edu> wrote:

> On 3/11/24 12:08 AM, Kerin Millar wrote:
> 
> > Speaking of which, to do both of these things has some interesting effects 
> > ...
> > 
> > $ z() { local -g a; unset -v a; a=123; echo "innermost: $a"; }; unset -v a; 
> > x; declare -p a
> > innermost: 123
> > inner: 123
> > outer: 123
> > declare -- a
> > 
> > $ z() { local -g a; unset -v a; unset -v a; a=123; echo "innermost: $a"; }; 
> > unset -v a; x; declare -p a
> > innermost: 123
> > inner: 123
> > outer: 123
> > declare -- a="123"
> 
> These show the normal effects of unset combined with dynamic scoping. This
> ability to unset local variables in previous function scopes has been the
> subject of, um, spirited discussion.

Of course, you are right. I had confused myself into thinking that there was 
some curious interaction between -g and the scope peeling technique on display 
there but I just tested again and could observe no adverse interaction. I'll 
attribute that to only being half-awake at the time.

> 
> > $ x() { local a; y; local +g a; a=456; echo "outer: $a"; }; unset -v a; x; 
> > declare -p a
> > innermost: 123
> > inner: 123
> > outer: 456
> > declare -- a="123"
> 
> Let's assume the previous function definitions remain in effect here. In
> that case, z continues to unset the local variable definitions in previous
> scopes, but the `local +g a' is basically a no-op -- it implies not using
> the global scope for a, which is the same as not using the option at all.

Thanks. To confirm this, I replaced "local +g a" with "local a" and saw that it 
continued to have the same effect of preventing the assignment of 456 from 
happening in the global scope. It came as a minor surprise but does make sense, 
given the purpose of local.

-- 
Kerin Millar