Conditional regex

Dear Monks,

If I use the construct (?(condition)yes-pattern) in a regex, and the condition is the execute-perl-code construct (?{1}), i.e. the condition is always true, the output is as expected:

my $str = 'bxAybz';

while ( $str =~ /(?(?{1})(b[xyz]))/g ) {
    say 'yes';
    say $1;
}

--output:--
yes
bx
yes
bz
[download]

But when I use the condition (?{pos() % 2 == 0}), I expect the same output, but I don't get it:

my $str = 'bxAybz';

while ( $str =~ /(?(?{pos() % 2 == 0})(b[xyz]))/g ) {
    say 'yes';
    say $1;
}

--output:--
yes
bx
yes
Use of uninitialized value $1 in say at 2.pl line 9.

yes
bz
[download]

Three matches?

Also, I notice the x modifier doesn't work with a conditional regex:


my $str = 'bxAybz';

while ( $str =~ /
    (?
        (?{1})
        (b[xyz])
    )
/gx ) {
    say 'yes';
    say $1;
}


--output:--
Sequence (?
...) not recognized in regex; marked by <-- HERE in m/
    (?
 <-- HERE         (?{1})
        (b[xyz])
    )
/ at 2.pl line 12.

[download]

That feels like a bug.

Comment on Conditional regex Select or Download Code
Re: Conditional regex by Anonymous Monk on Feb 17, 2013 at 18:35 UTC
But when I use the condition (?{pos() % 2 == 0}), I expect the same output, but I don't get it: Employ Basic debugging checklist or add `use re 'debug';` to see what the regex engine is doing, and why your match fails then adjust your pos invocation Also, your code wont run as is -- come on :) use feature or whatever Also, I notice the x modifier doesn't work with a conditional pattern: It does , you just can't have space in condition, it has to be `(?(?{`, otherwise you couldn't distinguish between some constructs	[reply] [d/l]
Re^2: Conditional regex by Anonymous Monk on Feb 17, 2013 at 18:49 UTC
perlre says (?(condition)yes-pattern\|no-pattern) (?(condition)yes-pattern) Conditional expression. Matches yes-pattern if condition yields a true value, matches no-pattern otherwise. A missing pattern always matches.	[reply]
Re^2: Conditional regex by AnomalousMonk (Prior) on Feb 17, 2013 at 19:30 UTC
It does , you just can't have space ... otherwise you couldn't distinguish between some constructs Further to Anonymonk's point: This is common. Eliminating the extraneous space around `?` in the example below allows this pedestrian regex to compile and work as expected. `>perl -wMstrict -le "print 'match' if 'xxx' =~ m{ ( ? : x ) }xms; " Sequence (? ...) not recognized in regex...` [download]	[reply] [d/l] [select]


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Conditional regex

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