CODEWITHSUNDEEP
perlmatchingregex-lookarounds
Simone
Simone

Reputation: 21

Why does a positive lookahead lead to captures in my Perl regex?

I can't get why this code work:

$seq = 'GAGAGAGA';
my $regexp = '(?=((G[UCGA][GA]A)|(U[GA]CG)|(CUUG)))'; # zero width match
while ($seq =~ /$regexp/g){ # globally
     my $pos = pos($seq) + 1; # position of a zero width matching
     print "$1 position $pos\n";
}

I know this is a zero width match and it dosn't put the matched string in $&, but why does it put it in $1?

thank you!

Upvotes: 2

Views: 1094

Answers (2)

Greg Bacon
Greg Bacon

Reputation: 139531

Matches are captured in $1 because of all the internal parentheses. If you don't want capturing, then use

my $regexp = '(?=(?:(?:G[UCGA][GA]A)|(?:U[GA]CG)|(?:CUUG)))';

or even better

my $regexp = qr/(?=(?:(?:G[UCGA][GA]A)|(?:U[GA]CG)|(?:CUUG)))/;

From the perlre documentation:

  • (?:pattern)
  • (?imsx-imsx:pattern)

This is for clustering, not capturing; it groups subexpressions like (), but doesn't make backreferences as () does. So

@fields = split(/\b(?:a|b|c)\b/)

is like

@fields = split(/\b(a|b|c)\b/)

but doesn't spit out extra fields. It's also cheaper not to capture characters if you don't need to.

Any letters between ? and : act as flags modifiers as with (?imsx-imsx). For example,

/(?s-i:more.*than).*million/i

is equivalent to the more verbose

/(?:(?s-i)more.*than).*million/i

Upvotes: 6

rjh
rjh

Reputation: 50294

Your regular expression contains a capture (...) which means the $1, $2, etc. variables will be populated with the results of those captures. This works in lookahead assertions too (although not lookbehind assertions, I believe).

As with all captures, if you rewrite as (?:...) then the contents will not go into a capture variable.

Upvotes: 2

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