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, $v) == pop %hash;
make sense to anyone except me?
... the only time it's useful
On Thu, 10 Feb 2005 08:59:04 -0800, David Storrs [EMAIL PROTECTED] wrote:
On Wed, Feb 09, 2005 at 05:13:56AM -0600, Rod Adams wrote:
Does
($k, $v) == pop %hash;
or
($k, $v) == %hash.pop;
make sense to anyone except me?
... the only time it's useful is
if you want to process
Does
($k, $v) == pop %hash;
or
($k, $v) == %hash.pop;
make sense to anyone except me?
Since we now have an explicit concept of pairs, one could consider a
hash to be nothing but an unordered (but well indexed) list of pairs.
So, C pop %hash would be a lot like C each , except, of course
Rod Adams wrote:
Does
($k, $v) == pop %hash;
or
($k, $v) == %hash.pop;
make sense to anyone except me?
Makes sense to me. Although I would be more inclined to think of pop as
returning a pair - but does a pair in list context turn into a list of
key, value? If so then the above makes lots
Matthew Walton [EMAIL PROTECTED] writes:
Rod Adams wrote:
Does
($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 into a list
