perl - 为什么正面前瞻会导致在我的 Perl 正则表达式中进行捕获？

Question

我不明白为什么这段代码有效:

$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";
}

我知道这是一个零宽度匹配，它不会将匹配的字符串放在 $& 中，但为什么将它放在 $1 中？

谢谢!

Answer 1

由于所有内部括号，匹配项被捕获在 $1 中。如果您不想捕获，请使用

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

甚至更好

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

来自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