On Sat Jul 07 22:23:16 2012, thoughtstream wrote:
> Father Chrysostomos asked:
>
> > What I am really trying to find out is when the subroutine is actually
> > cloned,
>
> Yes. It is supposed to be (or at least must *appear* to be),
> and currently is (or appears to be) in Rakudo.
I said when, not whether. :-)
> > and whether there can be multiple clones within a single call of
> > the enclosing sub.
>
> Yes. For example, a lexical sub might be declared in a loop inside the
> enclosing sub, in which case it should produce multiple instances, one
> per iteration.
>
> For example, this:
>
> sub outer_sub () {
> for (1..3) {
> state $call_num = 1;
> my sub inner_sub {
> state $inner_state = (1..100).pick; # i.e. random number
> say " [call {$call_num++}] \$inner_state =
$inner_state";
> }
>
> say "\nsub id: ", &inner_sub.id;
> inner_sub();
> inner_sub();
> }
> }
>
> outer_sub();
>
> produces:
>
> sub id: -4628941774842748435
> [call 1] $inner_state = 89
> [call 2] $inner_state = 89
>
> sub id: -4628941774848253711
> [call 3] $inner_state = 16
> [call 4] $inner_state = 16
>
> sub id: -4628941774839825925
> [call 5] $inner_state = 26
> [call 6] $inner_state = 26
>
> under Rakudo
Thank you.
Does Perl 6 have an equivalent to this?
my $x;
for $x(1..10) {}
In this case, the loop reuses the $x slot in the pad, making the
existing $x name an alias to a different scalar. But, due to the way
Perl 5 closures work, it is, for all tents and porpoises, the same as this:
my $x;
for my $x(1..10) {}
> BTW, Both the above "yes" answers are consistent with (and can be
> inferred from) the previous explanation that:
>
> my sub foo { whatever() }
>
> is just a syntactic convenience for:
>
> my &foo := sub { whatever() }
Except that my sub foo happens upon block entry, right?
>
> HTH,
It does, but I am still trying to wrap my head around the fundamental
difference between 5 and 6 with regard to closures.
In Perl 5, $] in a piece of code is bound to *], not $], so it sees
changes made by local($]) (which actually puts a completely new scalar
in the *]{SCALAR} slot). But ‘my $x; sub { $x }’ is bound, not to the
$x slot in the outer block/sub/file, but to the actual scalar itself.
It seems that Perl 6 closures close over the slot, not the
scalar/array/etc. Is that right?
Anyway, I think I know how to implement this now.
The first time a ‘my’ sub is referenced or called, it is cloned. The
clone is stored for reuse, but that storage is localised to the current
block.
--
Father Chrysostomos
