On Sun, Mar 10, 2024 at 04:01:10PM -0400, Lawrence Velázquez wrote:
> Basically, without an assignment, "local -g" does nothing.
Well, the original purpose of -g was to create variables, especially
associative arrays, at the global scope from inside a function.
I think this thread has been asking about a completely different
application, namely to operate upon a global scope variable from
within a function where a non-global scope variable is shadowing the
global one.
(As far as I know, there isn't any sensible way to do that in bash.)
Here it is in action. "local -g" (or "declare -g") without an assignment
in the same command definitely does things.
hobbit:~$ f() { declare -g var; var=in_f; }
hobbit:~$ unset -v var; f; declare -p var
declare -- var="in_f"
I think the key to understanding is that while "local -g var" creates
a variable at the global scope, any references to "var" within the
function still use the standard dynamic scoping rules. They won't
necessarily *see* the global variable, if there's another one at a
more localized scope.
