Since afaict this is not specced, I'll hand that over to p6l.
Eric Hodges (via RT) wrote:
use v6;
rule test {test};
test ~~ /test/;
say '$/.keys = ', $/.keys.perl;
say '%($/).keys = ', %($/).keys.perl;
# outputs
# $/.keys = []
# %($/).keys = [test]
Same could be said for .values
Jon Lang wrote:
I stand corrected. That said: with the eigenstates method now
private, it is now quite difficult to get a list of the eigenstates of
the above expression.
I thought about that a bit, and I think eigenstates are not hard to
extract (which somehow makes the privateness of
On Sat, Mar 28, 2009 at 10:39:01AM -0300, Daniel Ruoso wrote:
That happens because $pa and $pb are a singular value, and that's how
junctions work... The blackjack program is an example for sets, not
junctions.
Now, what are junctions good for? They're good for situation where it's
Richard Hainsworth conjectured:
1) Is the following true for an any junction?
any( ... , any('foo','bar')) === any(...,'foo','bar')
If yes, then
if an 'any' junction is contained in an outer 'any', the inner 'any' can be
factored out?
Yes. More precisely, an 'any' that is directly nested
On Sun, Mar 29, 2009 at 1:18 PM, John Macdonald j...@perlwolf.com wrote:
On Sat, Mar 28, 2009 at 10:39:01AM -0300, Daniel Ruoso wrote:
That happens because $pa and $pb are a singular value, and that's how
junctions work... The blackjack program is an example for sets, not
junctions.
Now,
Moritz Lenz wrote:
Since afaict this is not specced, I'll hand that over to p6l.
Eric Hodges (via RT) wrote:
use v6;
rule test {test};
test ~~ /test/;
say '$/.keys = ', $/.keys.perl;
say '%($/).keys = ', %($/).keys.perl;
# outputs
# $/.keys = []
# %($/).keys = [test]
Same could be
What I see here is that there is a tendency to want to think about,
and operate on, the eigenstates as a Set, but this seems to destroy
the single value impersonation of the Junction.
Further, if one ever calls .!eigenstates() on a Junction, then you
have really bollox'd your code up, as
