At 8:27 AM -0700 9/19/02, Brent Dax wrote:
>Dan Sugalski:
># Sort of, yes.
>#
># Basically the behaviour of hyper-operated operators is delegated via
> ^
>Spending time in England lately? ;^)
Why, yes, actually. :-P But I've been using Pompous English Spelling for years.
Dan Sugalski:
# Sort of, yes.
#
# Basically the behaviour of hyper-operated operators is delegated via
^
Spending time in England lately? ;^)
# multimethod dispatch to the hyper-operator functions. By default the
Well, yeah. But that doesn't really answer my question
At 12:04 AM -0700 9/18/02, Brent Dax wrote:
>The Apocalypse on operators says that if one of the operands of a
>hyperoperator is a scalar, then that scalar is (nominally) treated as an
>array of copies of that scalar. In other words:
>
> my $foo=1;
> my @bar=(2, 3, 4);
>
> my @b
The Apocalypse on operators says that if one of the operands of a
hyperoperator is a scalar, then that scalar is (nominally) treated as an
array of copies of that scalar. In other words:
my $foo=1;
my @bar=(2, 3, 4);
my @baz=$foo ^+ @bar; # @baz=(3, 4, 5)
Does
