Re: A regex that only matches at offset that are multiples of a given N?

[7stud]

in reply to A regex that only matches at offset that are multiples of a given N?

Help!

If I use the construct (?(condition)yes-pattern) in a regex, and the condition is (?{1}), i.e. 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

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

Three matches?

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

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.

[BrowserUk]

First, I'm not the right person to be asking this question of. Whatever regex expertise I once had is well out of date. There is a whole bunch of stuff I've never done anyting with. However, this is my interpretation of the decidedly unclear documentation:

Three matches?

The (what appears to be called) zero-length switch assertion, appears to succeed, whenever the condition part succeeds; regardless of whether the yes pattern (or no pattern, if present) succeed.

So with 6 characters in your string, and a condition that restricts matching to every second character, the overall match succeeds 3 times, even if the yes pattern only matches at 2 of those positions. Hence your output.

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

It appears that you are breaking up an indivisible token with the whitespace you've used. This works:

C:\test>perl -M5.010 -w
my $str = 'bxAybz';

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

[tobyink]

/(?(?{pos() % 2 == 0})(b[xyz]))/g

I think you want to explicitly back-track in the no condition...

/(?(?{pos() % 2 == 0})(b[xyz])(*FAIL))/g