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Postings from January 2012
On Mon Jan 17 08:49:39 2005, philippe.verdret@xps-pro.com wrote:
>
> This is a bug report for perl from pverdret@xps-pro.com,
> generated with the help of perlbug 1.35 running under perl v5.8.4.
>
> In a capture a reference to this capture is perfectly possible:
>
> "baaa" =~ (\1|a)+
>
> the result of this match is b<aaa>.
When I try out this code (or any of the subsequent examples) *as is*, I
get syntax errors.
#####
$ perl -e '"baaa" =~ (\1|a)+'
syntax error at -e line 1, at EOF
Execution of -e aborted due to compilation errors.
#####
But when I modify each of the code samples to make a proper regular
expression, I do not get the output the OP described:
> If I replace the back-reference \1 by a conditional expression like
> this:
>
> "baaa" =~ ((?(1)a|a))+
>
> the result is the same. The tested condition is
> independant of the capture content so if i change the regex like this:
>
> "baaa" =~ ((?(1)a|b))+
>
> the result should be "<baaaa>". But the matched string is <b>aaaa.
>
> By introducing a (?{}) behind "b" the result is now right:
>
> "baaa" =~ ((?(1)a|b(?{})))+
>
> use re 'debug' indicates that the empty code block changes the
> generated opcodes.
>
> Philippe Verdret
>
#####
$ perl -e '"baaa" =~ m/(\1|a)+/;print "$1\n"'
a
$ perl -e '"baaa" =~ m/((?(1)a|a))+/;print "$1\n"'
a
$ perl -e '"baaa" =~ m/"baaa" =~ ((?(1)a|b))+/;print "$1\n"'
$ perl -e '"baaa" =~ m/"baaa" =~ ((?(1)a|b(?{})))+/;print "$1\n"'
#####
So either I'm misinterpreting the OP, or his original premise was
incorrect, or Perl has changed since this was posted seven years ago.
Which?
Thank you very much.
Jim Keenan
---
via perlbug: queue: perl5 status: new
https://rt.perl.org:443/rt3/Ticket/Display.html?id=33820
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