Larry Wall wrote:
>
> : In summary: assuming Perl 6 allows user-defined while-ish structures, how
> : would it be done?
>
> I think the secret is to allow easy attachment of regex rules to sub
> and parameter declarations. There's little point in re-inventing
> regex syntax using declarations.
In a message dated Thu, 29 Aug 2002, Damian Conway writes:
> And, of course, the C property would smart-match its value
> against the corrresponding argument, so one could also code optimized
> variants like:
>
> sub repeat is multi ($desc is valued(1), &body) {
> body(1);
>
[apologies to anyone who received this twice... the bonehead at the keyboard
left the subject blank. ]
> This week I was fortunate enough to hear Damian speak twice, once on
> everything and once on Perl6. Damian, it was tremendous of you to come
> and speak to us in London - thank-you very muc
Damian Conway wrote:
> rule expr1 {
> { m:cont/@operators/ or fail }
> }
>
> Backtracking would just step back over the rule as if it were atomic
> (or followed by a colon).
Ok, thanks. (The "followed by a colon" is just to explain the behavior,
right? It's illegal to follow a
[NOTE: BCCing off-list to protect private email addresses]
On Fri, 2002-08-30 at 09:07, Ken Fox wrote:
> Does the following example backtrack into ?
>
>rule foo { b+ }
>rule bar { a b }
This was the bit that got me on-board. I did not see the need for
backtracking into rules until he
On Fri, 30 Aug 2002, Ken Fox wrote:
: Ok, thanks. (The "followed by a colon" is just to explain the behavior,
: right? It's illegal to follow a code block with a colon, isn't it?)
I don't see why it should be illegal--it could be useful if the closure
has played continuation games of some sort to
Larry Wall wrote:
> On Fri, 30 Aug 2002, Ken Fox wrote:
> : Ok, thanks. (The "followed by a colon" is just to explain the behavior,
> : right? It's illegal to follow a code block with a colon, isn't it?)
>
> I don't see why it should be illegal--it could be useful if the closure
> has played cont
On Fri, 30 Aug 2002, Ken Fox wrote:
: Apoc 5 has "It is an error to use : on any atom that does no
: backtracking." Code blocks don't backtrack (at least that's what
: I understood Damian to say).
Code blocks don't backtrack *by default*. But you can do anything
in a closure.
: Are zero width a
Larry Wall wrote:
> There's a famous book called "Golf is Not a Game of Perfect".
Well now I'm *totally* confused. I looked that up on Amazon
and it has something to do with clubs and grass and stuff. That's
completely different than what I thought golfing was. ;)
Seriously, though. I have a pos
Exegesis 4 says
When the subroutine dispatch mechanism detects one or more pairs as
arguments to a subroutine with named parameters, it examines the keys of
the pairs and binds their values to the correspondingly named parameters
-- no matter what order the paired arguments originally app
In Damian's excellent perl6 talk, I think he said that by default a hash
in list context will return a list of pairs. Hence this
@array = %hash
for %hash with n keys would give an array of n elements, all pairs.
If you want the perl5 tradition of a list alternating key,value,key,value...
you'
Nicholas Clark wrote:
[...]
> And what happens if I write
>
> %hash4 = ("Something", "mixing", pairs => and, "scalars");
1 23 4 5
Perl5 says "Odd number of elements in hash assignment at -e line 1."
And Perl6 should, too.
IMHO, your example isn't
[EMAIL PROTECTED] (Steffen Mueller) writes:
> > %hash4 = ("Something", "mixing", pairs => and, "scalars");
>1 23 4 5
> Perl5 says "Odd number of elements in hash assignment at -e line 1."
> And Perl6 should, too.
Except that a pair is a single th
Steffen Mueller
> > %hash4 = ("Something", "mixing", pairs => and, "scalars");
>1 23 4 5
> Perl5 says "Odd number of elements in hash assignment at -e line 1."
> And Perl6 should, too.
Hmm, I rather like the idea of thinking of a %foo variable as
Piers Cawley wrote:
> Maybe we should just say 'sod it' and implement the entire Smalltalk
> Collection hierarchy and have done with it? Sets, bags, hashes
> (dictionaries for the Smalltalker), whatever, all have their uses...
I'm not sure if you were being facetious, but I do think all the
funct
Is C<\n> going to be a rule (e.g. C<< >>) or is it implicitly
translated to:
<[\x0a\x0d...]>+
If it's the latter, then what does this do?
\n?
Do I get
[<[\x0a\x0d...]>+]?
Or do I get
<[\x0a\x0d...]>+?
If the former (which I assume is the case), how do I get
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