Re: "unset var" pops var off variable stack instead of unsetting it

[Stephane Chazelas]

2017-03-19 18:05:19 -0400, Chet Ramey:
> On 3/19/17 5:51 PM, Stephane Chazelas wrote:
> 
> > On comp.unix.shell ot http://unix.stackexchange.com, I've posted
> > many articles describing how to do splitting in POSIX-like
> > shells:
> > 
> > ( # subshell for local scope
> >   unset -v  IFS # restore default splitting behaviour
> >   set -o noglob # disable globbing
> >   cmd -- $var   # split+glob with default IFS and glob disabled
> > )
> > 
> > I'm now considering adding a note along the lines of:
> > 
> >   "Beware that with current versions of bash, pdksh and yash,
> >   the above may not work if used in scripts that otherwise use
> >   typeset/declare/local on $IFS or call a function with
> >   `IFS=... my-function' (or IFS=... eval... or IFS=...
> >   source...)"
> 
> You can, of course, do whatever you want.  You might want to read my
> message from yesterday about what happens when you do that, or look
> at the following examples, after which you may decide that the situation
> is not as dire.
> 
> $ cat x2
> function foo
> {
> (
>       unset -v IFS
>       recho "${IFS-unset}"
> )
> }
> 
> IFS=':|'
> foo
> echo after IFS = "$IFS"
> $ ../bash-4.4-patched/bash ./x2
> argv[1] = <unset>
> after IFS = :|

Yes, that one is  fine but it is not the issue that is being
discussed here. There's no variable to pop off a stack above.

the issue is when that "foo" function is called in a context
where IFS had been declared locally. Like in:

IFS=1
function example {
  typeset IFS=2
  foo
}

Where "foo" would output "1", because then "unset -v IFS" would
*not* have unset IFS but instead would have restored the value
it had before the "typeset" (in that case, the global scope).

-- 
Stephane