[perl #77616] ~ operator in regexp reverts capture order, but it should not

2014-10-09 Thread Christian Bartolomaeus via RT
Thanks for the feedback. I added a fudged test to S05-metachars/tilde.t with the following commit: https://github.com/perl6/roast/commit/f15d9aed26

[perl #77616] ~ operator in regexp reverts capture order, but it should not

2014-10-08 Thread Christian Bartolomaeus via RT
Current behaviour: > say so "abc" ~~ /a ~ (c) (b)/; say $0, $1 True 「b」 「c」 As I understand S05 the ~ operator basically rewrites the above regex to /a (b) (c)/ and only then matching and capturing happens. So the question seems to be, whether numbering of subpatterns should be done before the

[perl #77616] ~ operator in regexp reverts capture order, but it should not

2012-10-21 Thread Will Coleda via RT
On Thu Aug 25 03:17:17 2011, bbkr wrote: > [12:14] nom: say so "abc" ~~ /a ~ (c) (b)/; say $0 # this is > weird. in rakudo it captured incorectly ( #77616 ) in nom it does not > capture at all. > [12:14] nom b0da69: OUTPUT«Bool::True␤Any()␤» Current nom behavior: > say so "abc" ~~ /a ~ (c) (b)/

[perl #77616] ~ operator in regexp reverts capture order, but it should not

2010-09-01 Thread via RT
# New Ticket Created by Paweł Pabian # Please include the string: [perl #77616] # in the subject line of all future correspondence about this issue. # http://rt.perl.org/rt3/Ticket/Display.html?id=77616 > [17:29] rakudo: say so "abc" ~~ /a ~ (c) (b)/; say $0 [17:29] rakudo dc9900: OUTPUT«1␤