Larry Wall wrote:
Presumably
$obj does (A,B,C)
could also be made to work even with non-associative does.
Right.
Note that you _do_ want to be able to do something to the effect of
ordered composition in a single statement, though:
role A {
...
method m { ... }
}
Author: larry
Date: Fri Oct 20 17:01:15 2006
New Revision: 13252
Modified:
doc/trunk/design/syn/S12.pod
Log:
"does" operator is non-associative according to S03. Leave it that way for now.
Modified: doc/trunk/design/syn/S12.pod
===
On Fri, Oct 20, 2006 at 04:47:04PM -0700, Larry Wall wrote:
: For now the conservative thing is probably that we should just leave
: "does" as non-associative and make you write
:
: $obj.also_does(A,B,C)
:
: or some such if you want to get fancy.
Presumably
$obj does (A,B,C)
could also
On Thu, Oct 19, 2006 at 03:31:18PM -0700, Jonathan Lang wrote:
: Larry Wall wrote:
: >Though actually, now that I think about it, the cascaded notation
: >in S12 is illegal according to S03, since "does" is classified as
: >non-chaining, which implies non-associative.
:
: Wait a minute. Isn't "ch
Larry Wall wrote:
Anyway, I think the type constructors need to avoid overloading the
logic operators. Perl 6 is an operator-rich language because that
contributes to long-term clarity despite the short-term confusion.
OK. My main dog in this race is the idea of defining new roles
through the
On Fri, Oct 20, 2006 at 09:10:12AM -0700, Jonathan Lang wrote:
: TSa wrote:
: >Here is yet another idea to go with the two lattice operations:
: >
: > /\ meet also: infimum, intersection, glb (greatest lower bound)
: > \/ join also: supremum, union,lub (least upper bound)
:
: I ha
TSa wrote:
Here is yet another idea to go with the two lattice operations:
/\ meet also: infimum, intersection, glb (greatest lower bound)
\/ join also: supremum, union,lub (least upper bound)
I have to admit: if it turns out that '&' and '|' can't be used for
'intersection'
TSa wrote:
Jonathan Lang wrote:
> In short, R3 isn't neccessarily a subset of A; it's a superset of A &
> B. In a partial ordering graph, there's no reliable ordering between
> R3 and A.
>
> The standard syntax for creating roles can't reliably produce a subset
> of an existing role, because it
HaloO,
I wrote:
Yes, but I was conjecturing that the additions to A&B are pushed
down to A and B such that their intension sets remain strict supersets
of A&B.
Think of the Complex example that might read
role Complex does Num & !Comparable
{
method im { return 0; }
method re
HaloO,
I wrote:
In fact if we decide to specify a role combination syntax then it
should be the same everywhere. That means in a signature A|B would
require a more specific type and pure A or B wouldn't be admissible.
To get the old meaning of | you have to write A&B or perhaps the
juxtaposition
How does one do this:
http://www.davidnaylor.co.uk/archives/2006/10/19/threaded-data-collection-with-python-including-examples/
in perl 6? Assumin get_feed_list, get_feed_contents, parse_feed, and
store_feed_items are handled by modules like LWP and XML::Parser.
Will there be something native a
HaloO,
Jonathan Lang wrote:
In short, R3 isn't neccessarily a subset of A; it's a superset of A &
B. In a partial ordering graph, there's no reliable ordering between
R3 and A.
The standard syntax for creating roles can't reliably produce a subset
of an existing role, because it always allows
TSa wrote:
Jonathan Lang wrote:
>role R3 does A & B { ... }
>R3.does(A) # false: R3 can't neccessarily do everything that A can.
>A.does(R3) # false: A can't neccessarily do everything that R3 can.
That last one should be true. Role R3 contains the things that A and B
have in common.
Jonathan Lang skribis 2006-10-19 18:27 (-0700):
> Let's say that I want
>
>$expression?;
>
> to mean the same thing as the statement
>
>$_ = $expression;
>
> That is, any statement that ends with a '?;' instead of a ';'
> evaluates in scalar context instead of void context and stores th
HaloO,
Jonathan Lang wrote:
role R3 does A & B { ... }
R3.does(A) # false: R3 can't neccessarily do everything that A can.
A.does(R3) # false: A can't neccessarily do everything that R3 can.
That last one should be true. Role R3 contains the things that A and B
have in common. Hence e
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