Capturing separator (like /something % (.+?)/ ) spews out backtracking

[p6rt]

Migrated from rt.perl.org#125285 (status was 'resolved')

Searchable as RT125285$

[p6rt]

From @AlexDaniel

This command​:
perl6 -e "say 'rule1 foo rule2 bar' ~~ /^ ( 'rule1' || 'rule2' )* %% (.+?)
$/"

Outputs this​:
｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ ｣
1 => ｢ f｣
1 => ｢ fo｣
1 => ｢ foo｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ ｣
1 => ｢ b｣
1 => ｢ ba｣
1 => ｢ bar｣

I can't see any reason why these partial matches should appear in the match
object. I expected this​:

｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ bar｣

[p6rt]

From @AlexDaniel

I works this way though​:
perl6 -e "say 'rule1 foo rule2 bar' ~~ /^ ( 'rule1' || 'rule2' )* %%
$<baz>=[.+?] $/"

｢rule1 foo rule2 bar｣
0 => ｢rule1｣
baz => ｢ foo ｣
0 => ｢rule2｣
baz => ｢ bar｣

Change [] to () and it will break the same way.

On Sat, May 30, 2015 at 3​:05 AM, perl6 via RT <perl6-bugs-followup@​perl.org>
wrote​:

Greetings,

This message has been automatically generated in response to the
creation of a trouble ticket regarding​:
"Capturing separator (like /something % (.+?)/ ) spews out
backtracking",
a summary of which appears below.

There is no need to reply to this message right now. Your ticket has been
assigned an ID of [perl #​125285].

Please include the string​:

     \[perl #​125285\]

in the subject line of all future correspondence about this issue. To do
so,
you may reply to this message.

                    Thank you,
                    perl6\-bugs\-followup@​perl\.org

-------------------------------------------------------------------------
This command​:
perl6 -e "say 'rule1 foo rule2 bar' ~~ /^ ( 'rule1' || 'rule2' )* %% (.+?)
$/"

Outputs this​:
｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ ｣
1 => ｢ f｣
1 => ｢ fo｣
1 => ｢ foo｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ ｣
1 => ｢ b｣
1 => ｢ ba｣
1 => ｢ bar｣

I can't see any reason why these partial matches should appear in the match
object. I expected this​:

｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ bar｣

[p6rt]

From @labster

[7​:31pm] AlexDaniel​: m​: grammar MyGrammar { regex TOP { ^ [ <foo> <bar> ]+ $ }; regex foo { 'ZZZ' }; regex bar { .*? }; }; my $data = 'ZZZ 123 ZZZ 456'; say MyGrammar.parse($data);
[7​:31pm] camelia​: rakudo-moar c2a57e​: OUTPUT«｢ZZZ 123 ZZZ 456｣␤ foo => ｢ZZZ｣␤ bar => ｢｣␤ bar => ｢ ｣␤ bar => ｢ 1｣␤ bar => ｢ 12｣␤ bar => ｢ 123｣␤ bar => ｢ 123 ｣␤ foo => ｢ZZZ｣␤ bar => ｢｣␤ bar => ｢ ｣␤ bar => ｢ 4｣␤ bar => ｢ 45｣␤ …»
[7​:31pm] AlexDaniel​: now this is a problem
...
[7​:58pm] AlexDaniel​: m​: grammar MyGrammar { regex TOP { ^ [ <foo> $<bar>=.*? ]+ $ }; regex foo { 'ZZZ' }; }; my $data = 'ZZZ 123 ZZZ 456'; say MyGrammar.parse($data);
[7​:59pm] camelia​: rakudo-moar c2a57e​: OUTPUT«｢ZZZ 123 ZZZ 456｣␤ foo => ｢ZZZ｣␤ bar => ｢ 123 ｣␤ foo => ｢ZZZ｣␤ bar => ｢ 456｣␤»

So if you define <bar> inline, you get a single result, but if it's a regex you get multiple, backtracking(?) results.

On Sat Jun 06 14​:54​:18 2015, alex.jakimenko@​gmail.com wrote​:

I works this way though​:
perl6 -e "say 'rule1 foo rule2 bar' ~~ /^ ( 'rule1' || 'rule2' )* %%
$&lt;baz&gt;=[.+?] $/"

｢rule1 foo rule2 bar｣
0 => ｢rule1｣
baz => ｢ foo ｣
0 => ｢rule2｣
baz => ｢ bar｣

Change [] to () and it will break the same way.

On Sat, May 30, 2015 at 3​:05 AM, perl6 via RT <perl6-bugs-
followup@​perl.org>
wrote​:

Greetings,

This message has been automatically generated in response to the
creation of a trouble ticket regarding​:
"Capturing separator (like /something % (.+?)/ ) spews out
backtracking",
a summary of which appears below.

There is no need to reply to this message right now. Your ticket has
been
assigned an ID of [perl #​125285].

Please include the string​:

[perl #​125285]

in the subject line of all future correspondence about this issue. To
do
so,
you may reply to this message.

Thank you,
perl6-bugs-followup@​perl.org

-------------------------------------------------------------------------
This command​:
perl6 -e "say 'rule1 foo rule2 bar' ~~ /^ ( 'rule1' || 'rule2' )* %%
(.+?)
$/"

Outputs this​:
｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ ｣
1 => ｢ f｣
1 => ｢ fo｣
1 => ｢ foo｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ ｣
1 => ｢ b｣
1 => ｢ ba｣
1 => ｢ bar｣

I can't see any reason why these partial matches should appear in the
match
object. I expected this​:

｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ bar｣

[p6rt]

From @jnthn

On Fri May 29 17​:05​:32 2015, alex.jakimenko@​gmail.com wrote​:

This command​:
perl6 -e "say 'rule1 foo rule2 bar' ~~ /^ ( 'rule1' || 'rule2' )* %% (.+?)
$/"

Outputs this​:
｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ ｣
1 => ｢ f｣
1 => ｢ fo｣
1 => ｢ foo｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ ｣
1 => ｢ b｣
1 => ｢ ba｣
1 => ｢ bar｣

I can't see any reason why these partial matches should appear in the match
object. I expected this​:

｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ bar｣

This one took some figuring out... Anyway, fixed now​:

$ perl6 -e "say 'rule1 foo rule2 bar' ~~ /^ ( 'rule1' || 'rule2' )* %% (.+?) $/"
｢rule1 foo rule2 bar｣
0 => ｢rule1｣
1 => ｢ foo ｣
0 => ｢rule2｣
1 => ｢ bar｣

Tests in S05-match/capturing-contexts.t.

/jnthn

[p6rt]

The RT System itself - Status changed from 'new' to 'open'

[p6rt]

@jnthn - Status changed from 'open' to 'resolved'