On 2/26/18 4:31 AM, Robert Elz wrote:
> Date: Mon, 12 Feb 2018 09:26:37 -0500
> From: Chet Ramey <chet.ra...@case.edu>
> Message-ID: <790ade74-690f-541c-9ab4-635991744...@case.edu>
>
> | This is bash's dynamic scoping. The visibility of a local variable is
> | restricted to a function and its children, and `unset' removes the
> | currently-visible instance. Removing such an instance can `unconver' an
> | instance in a previous scope.
>
> Frankly this is brain dead, unset should not be unlocal (or something equiv)
>
> eg: if I have a func
>
> myfunc() {
> local IFS
> unset IFS
> # do some code
> }
>
> the very last thing that I want is for the global IFS to apply.
It doesn't. Run the following script:
func()
{
local var=$'a\tb\tc'
typeset IFS=' '
echo ${FUNCNAME}: before unset: $var
unset IFS
echo ${IFS:-null or unset}
echo ${FUNCNAME}: after unset: $var
}
IFS='%'
declare -p IFS
func
declare -p IFS
You'll see that the first expansion of `$var' uses the local value of IFS,
the second expansion uses the default value of $' \t\n', and the global
value doesn't change (or get unset) outside the function.
The objection was that the global or previous-scope value didn't get unset
when using the `unset' builtin; only in the local scope was it unset.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU c...@case.edu http://tiswww.cwru.edu/~chet/
