On 11/3/05, Luke Palmer <[EMAIL PROTECTED]> wrote:
> If Foo2 were a role (that is, if it obeys the role relation above),
> then the only thing bar2() could do would be to take some side-effect
> action and then return the same object it was passed. Here's a proof:
>
> Given ^T $x where Foo{^T}
On 11/3/05, TSa <[EMAIL PROTECTED]> wrote:
> So if A <: B, I would expect Role{B} <: Role{A} and Factory{A} <: Factory{B}
> on the following rational.
Well, it's possible that I'm abusing the terms, since I first heard
the terms from you and inferred what they meant.
However, there is a problem i
HaloO,
I don't understand why theory.pod states that roles are covariant, unary
theories and factories are contravariant. I would expect the opposite from
the requirement that all functions in roles only take the topic type while
function in factories only return the topic type.
So if A <: B, I
HaloO,
Rob Kinyon wrote:
On Nov 2, 2005, at 9:02 PM, Jonathan Lang wrote:
Let's say you have this:
role A {method foo() { code1; } }
role B {method foo() { code2; } }
I think, A and B might just be aliases to the *identical* structural type
because the only constraint that both roles impo
HaloO,
Luke Palmer wrote:
On 10/29/05, Damian Conway <[EMAIL PROTECTED]> wrote:
So we need a mechanism that is externally (i.e. from a class interface
point-of-view) a subroutine, but internally has the features of a method (i.e.
has an invocant). Since it's externally sub-like but internally
> On Nov 2, 2005, at 9:02 PM, Jonathan Lang wrote:
> > Let's say you have this:
> >
> > role A {method foo() { code1; } }
> > role B {method foo() { code2; } }
> > role C does A does B {
> > method foo() { A::foo(); }
> > method bar() { B::foo(); }
> > }
> >
> > Should the following
On Nov 2, 2005, at 9:02 PM, Jonathan Lang wrote:
Let's say you have this:
role A {method foo() { code1; } }
role B {method foo() { code2; } }
role C does A does B {
method foo() { A::foo(); }
method bar() { B::foo(); }
}
Should the following be valid?
role D does C { method
On 11/3/05, Michele Dondi <[EMAIL PROTECTED]> wrote:
> On Wed, 2 Nov 2005, Ruud H.G. van Tol wrote:
>
> >> http://www.nntp.perl.org/group/perl.perl6.language/17556
> >
> > I understand that Perl6 allows blocks with changed/enhanced syntax, so
> > it is or will become possible (to add it) as if it w
On Thu, 3 Nov 2005, Sam Vilain wrote:
That being said, there are probably other more pressing reasons that ops
should not accept $_ as default; I would guess, for a start, it makes
determining semantics very difficult. Does ++; mean &postfix:<++> or
&prefix:<++> ?
If we had it, I think we wo
On Wed, 2 Nov 2005, Ruud H.G. van Tol wrote:
http://www.nntp.perl.org/group/perl.perl6.language/17556
I understand that Perl6 allows blocks with changed/enhanced syntax, so
it is or will become possible (to add it) as if it was in the core
language.
Do I understand that right? Something as sim
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