On May 23, 2009 11:26:16 pm John M. Dlugosz wrote:
> > From whence did it get its Item container?
>
> OK, my brain made a wrong turn some time on Tuesday.
>
> Let me review the basics.
>
> From S02:
> |$x| may be bound to any object, including any object that can be bound
>
> to any other sigil.
>
> Perl variables have two associated types: their "value type" and their
> "implementation type".
> Value type = "of", implementation type is the type of the container itself.
>
> my $spot is Scalar; # this is the default
>
I think (that I may be getting out of my depth here) that the implementation
type refers to the container at a deeper level. That is, Scalar is a Perl
container and not a Java nor Ruby container, which allows having objects from
different languages (with their different semantics) in a single application.
>
> From S03:
>
> A new form of assignment is present in Perl 6, called /binding/, used in
> place of typeglob assignment. It is performed with the |:=| operator.
> Instead of replacing the value in a container like normal assignment, it
> replaces the container itself. For instance:
>
> my $x = 'Just Another';
> my $y := $x;
> $y = 'Perl Hacker';
>
At first, I thought that this was a aliasing mechanism, but now I worry that
this is how we have to assign non-Perl variables in a Perl application. That
is, the semantics of assignment differ depending the container type.
> From S12:
>
> Method calls on mutable scalars always go to the object contained in the
> scalar (autoboxing value types as necessary):
>
> $result = $object.doit();
> $length = "mystring".codes;
>
> Method calls on non-scalar variables just calls the |Array|, |Hash| or
>
> |Code| object bound to the variable:
>
> $elems = @array.elems;
> @keys = %hash.keys;
> $sig = &sub.signature;
>
> Use the prefix |VAR| macro on a scalar variable to get at its underlying
>
> |Scalar| object:
>
> if VAR($scalar).readonly {...}
>
> ||============================
>
> So, if you say
>
> my $x;
>
> then $x is bound to a newly-created Scalar (not worrying about
> conflating the name the Role and the concrete type at this point).
>
> my $x = 'Just Another';
>
> Now the "item assignment" operates on the container, storing the Str
> instance within it. $x is bound to a container, and the container
> contains (only) one Str instance.
>
Rephrasing: $x is a Perl container holding a Str object whose content
(value?) is 'Just Another'.
> Basically, I was thinking that scalar variables always are bound to item
> containers which contain the actual value. In fact, this was originally
> drilled into me: containers vs values! That variables always directly
> hold some kind of container. I suppose that's been changed at some
> point, perhaps long ago, but the synopses didn't dissuade me from that
> early belief.
>
I don't think it has been changed, I think that if you don't understand how
"tie" is implemented in Perl5 (which is referenced heavily in the synopsis),
then it is easy to miss the distinctions being made. (It doesn't help that
the Implementation type is in the middle of a bunch of "higher level" types).
>
> Anyway, as for your (Henry) example from Rakudo:
>
> my $x = 1, 2, 3;
>
> According to the synopses, this should be analogous to the previous
> example, where $x now is bound to an item that contains a Capture
> containing 3 Ints.
>
> Assuming Capture vs Array is simply out of date implementation, just
> focus on the parameter passing. $x is bound to a Scalar. But What I
> Mean is for the formal parameter @y to get bound to the item in the
> container, the list of Ints.
>
> ===> How does that happen?
>
The "List prefix precedence" section of Synopsis 3 says that it returns a list.
It goes on a bit about the "contortions" that must be done to do this.
>
> Now back to straightening out my misconceptions about scalars _always_
> holding item containers. If $x is bound to an Array, for example, the
> compiled code can't be doing the indirection innately.
>
> So it follows that the method forwarding is a property of the object
> that the method is originally called on. That is, the Scalar (and any
> "tied" item container implementations) are written to forward all
> methods to the contained object. The compiler sees $x.foo() and doesn't
> know anything about $x other than that it's some object, so it codes a
> message dispatch to it. If $x holds a normal object, the foo method
> gets called via the normal dispatching rules. But Scalar implements
> something that catches all methods and forwards them to the contained item.
>
> ===> Right?
>
Not quite (I think). As mentioned above, I think that the Scalar is the the
thing that knows how/which dispatch to use.
> That would imply that if a scalar happened to contain another scalar, e.g.
> my $x = Scalar.new;
> ($x is bound to a Scalar which contains a Scalar which contains undef)
> then any method called on $x would trigger the same behavior when the
> contained object gets it, and be forwarded all the way down, no matter
> how many Scalars I nested in this manner.
>
> So how does VAR work? It can't just wrap another level around the
> original object to cancel out the automatic indirection. It can't tell
> the compiler to generate different code, since the code knows nothing
> about this interesting property of scalars. Instead it must return a
> proxy that knows how to trigger the actual methods on the scalar,
> avoiding the forwarding behavior of normal method calls.
>
> ===> Is that right?
>
Relying on rakudo again:
> my $x; say $x.WHAT
Failure()
> my $x = Scalar.new; say $x.WHAT
Could not find non-existent sub Scalar
>
Regards,
Henry
> Thanks for your help everyone. I hope to give back just as much, once
> I've caught back up. In any case, what I learn I will document for all
> who come later.
>
> --John
--
Henry Baragar
Instantiated Software
416-907-8454 x42
