On Fri, Feb 11, 2005 at 05:33:29PM -0800, Ashley Winters wrote:
> On Thu, 10 Feb 2005 08:59:04 -0800, David Storrs <[EMAIL PROTECTED]> wrote:
> > On Wed, Feb 09, 2005 at 05:13:56AM -0600, Rod Adams wrote:
> > >
> > > ($k, $v) <== pop %hash;
> > > make sense to anyone except me?
> >
> > ... the onl
On Thu, 10 Feb 2005 08:59:04 -0800, David Storrs <[EMAIL PROTECTED]> wrote:
> On Wed, Feb 09, 2005 at 05:13:56AM -0600, Rod Adams wrote:
> >
> > Does
> >
> > ($k, $v) <== pop %hash;
> > or
> > ($k, $v) <== %hash.pop;
> >
> > make sense to anyone except me?
>
> ... the only time it's useful is
> if
On Wed, Feb 09, 2005 at 05:13:56AM -0600, Rod Adams wrote:
>
> Does
>
> ($k, $v) <== pop %hash;
> or
> ($k, $v) <== %hash.pop;
>
> make sense to anyone except me?
It's clear to me.
The only thing is that, right off the top of my head, I can't see
where it would be used. Since the order in whi
Matthew Walton <[EMAIL PROTECTED]> writes:
> Rod Adams wrote:
>> Does
>> ($k, $v) <== pop %hash;
>> or
>> ($k, $v) <== %hash.pop;
>> make sense to anyone except me?
>
> Makes sense to me. Although I would be more inclined to think of pop
> as returning a pair - but does a pair in list context turn
Rod Adams wrote:
Does
($k, $v) <== pop %hash;
or
($k, $v) <== %hash.pop;
make sense to anyone except me?
Makes sense to me. Although I would be more inclined to think of pop as
returning a pair - but does a pair in list context turn into a list of
key, value? If so then the above makes lots of se
Does
($k, $v) <== pop %hash;
or
($k, $v) <== %hash.pop;
make sense to anyone except me?
Since we now have an explicit concept of pairs, one could consider a
hash to be nothing but an unordered (but well indexed) list of pairs.
So, C<< pop %hash >> would be a lot like C<< each >>, except, of cours
