On May 27, 2009 01:52:58 pm Mark J. Reed wrote:
> On Wed, May 27, 2009 at 1:42 PM, Daniel Carrera
>
> <daniel.carr...@theingots.org> wrote:
> > sub postfix:<!> { [*] 1..$^n }
> > say 5!;
> >
> > WOW!! That *IS* cool. Can you explain to me how it works? I figured out
> > postfix: myself, but the rest is obscure to me.
>
> Key concepts:
>
> 1. placeholder variables. The ^ in $^n means it's a placeholder: no
> predeclaration required, and placeholders in an expression are
> assigned the passed-in arguments in serial order. (The sub could also
> have been written more traditionally as sub postfix:<!>($n) { [*]
> 1..$n } .)
>
> 2. the range operator .. : $x..$y for integers $x and $y generates,
> in list context, a list of the integers from $x to $y, inclusive.
>
> 3. the reduction meta-operator [...] : [OP](@list) collects the
> result of applying OP to the elements of the list in order. That is,
> assuming foo() is a binary sub, [foo](1,2,3,4) =
> foo(foo(foo(1,2),3),4). So [+](@list) generates a sum of the listed
> values, [*] generates their product, etc.
>
> So, given the argument to !:
It might have been implied, but you kind of missed the discussion about the
ability to add new operators (and how they relate to other operators).
To recap (slightly differently) there are several impressive features that are
easier to explain and can be part of the build up to the factorial example:
Placeholder variables: non-declared and descriptive
Ranger operators: not so impressive and inherited from perl5 (etc.)
The reduction meta-operator [...]: nice syntax and no need to provide a
identity/seed element
A side discussion of other interesting meta-operators
User defined operators: overloading and novel mechanism of declaring
precedence (looser, tighter, etc.)
Finally, a simple slide where the factorial definition takes up the whole slide
and let the audience have a few moments of silence to let them figure it out
and to start to really appreciate the expressiveness/impressiveness of perl6
Henry
>
> 1. create a list of integers from 1 to that value (1..$^n)
> 2. multiply them all together ([*])
>
> and of course a sub without an explicit return statement returns the
> value of the last expression.
>
> >> I do think captures are inherently impressive, but not easy to
> >> explain...
> >
> > Got a link?
> >
> > Daniel.
--
Henry Baragar
Instantiated Software
416-907-8454 x42
