On Sun, 17 Aug 2003, Benjamin Goldberg wrote:
> Michal Wallace wrote:
>
> > Uh-oh. I just went to implement "del x"
> > and there's no op to remove a variable
> > from a lexical pad! :)
>
> Why would you want to remove a variable from a lexical pad?
>
> Surely the "right thing to do" would be to create a new pad (scope),
> then add your 'x' variable which you plan to 'delete' in the future,
> then when you want to delete that variable, pop off that pad.
Hmm. Do you mean
if for stmt in block:
if stmt.type == undef:
flag_as_going_to_delet(stmt.varname)
So I can create a new pad when it's assigned?
It can't be a simple pop though, can it?
#!/usr/bin/perl
$x = "cat";
$y = "pop $x?";
undef $x;
print "x = $x \n";
print "y = $y \n";
Because here, $x has to be defined before $y, so
I'd have to delete the -2nd scope. Unless the
code was smart enough to work out that y is in
-2 and x is in -1...
In any case, what does it buy me? it seems like a
lot of work when there's already a delete_keyed
op, and leo just made an implementation for pads. :)
I guess in perl it's not bad, since that's the
whole point of undef, you can just $x = PerlUndef
it... But in python, this throws a NameError:
x = 1
del x
x
So the easiest thing to me is just to translate
del x as "remove this from the current pad".
Sincerely,
Michal J Wallace
Sabren Enterprises, Inc.
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