**{x,y} quantifier

2007-07-01 Thread Amir E. Aharoni
(I'm just studying the intricacies of Perl 6, so please correct me if i say something stupid or if this has already been discussed before.) I was looking for the Perl 6 equivalent of "aaa" =~ /a{1,3}/ and finally found that it's "aaa" ~~ /a**{1 .. 3}/ This looked rather weird, so i asked on I

Re: **{x,y} quantifier

2007-07-01 Thread Luke Palmer
On 7/1/07, Amir E. Aharoni <[EMAIL PROTECTED]> wrote: I got the reply that it is similar to exponentiation of variables in math: a ** 5 == a * a * a * a * a == a It makes sense after it is explained and i do like the rationalization of the range as a list-like range, instead of the comma, b

Re: **{x,y} quantifier

2007-07-01 Thread Amir E. Aharoni
> please correct me if > i say something stupid or if this has already been discussed before.) Another important loss if we were to go with <1..3> would be the ability to have runtime-dependent ranges; e.g.: / ($ntimes) x**{$ntimes} / That's exactly what i meant by "something stupid". T

Re: **{x,y} quantifier

2007-07-01 Thread Luke Palmer
On 7/1/07, Amir E. Aharoni <[EMAIL PROTECTED]> wrote: > > please correct me if > > i say something stupid or if this has already been discussed before.) > Another important loss if we were to go with <1..3> would be the > ability to have runtime-dependent ranges; e.g.: > > / ($ntimes) x**{$n

Named captures (was: **{x,y} quantifier)

2007-07-01 Thread Amir E. Aharoni
On 01/07/07, Luke Palmer <[EMAIL PROTECTED]> wrote: On 7/1/07, Amir E. Aharoni <[EMAIL PROTECTED]> wrote: > > > please correct me if > > > i say something stupid or if this has already been discussed before.) > > > Another important loss if we were to go with <1..3> would be the > > ability to ha

Re: Named captures (was: **{x,y} quantifier)

2007-07-01 Thread Luke Palmer
On 7/1/07, Amir E. Aharoni <[EMAIL PROTECTED]> wrote: On 01/07/07, Luke Palmer <[EMAIL PROTECTED]> wrote: > / $ntimes := (\d+) x**{$ntimes} / The examples of := usage in S05 seem to have notation such as this: $ := (\d+) Yes, that is correct. I've been away from the Perl 6 community

A problem about IPC::Open2

2007-07-01 Thread Liang He
Hello, I'm a beginner of the PERL language and I have troubles in understanding a PERL program these days. Actually I've been confused for a long time.I hope you can look into this program and point out the problem for me:) The program is in the attachment.It's a bit long , but the error occured

Re: A problem about IPC::Open2

2007-07-01 Thread Nir Pengas
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Re: A problem about IPC::Open2

2007-07-01 Thread Jonathan Lang
Admittedly, that wasn't particularly germane to the perl6 mailing list; but it did bring up an issue that's been bothering me for a while. I would like to see Perl6 handle the equivalent of IPC::Open2 using the same sort of syntax and semantics that it uses for sockets, by default. That is, I'd