Please explain the meaning of this regular expression and what groups the expression will generate?
$string =~ m/^(\d*)(?: \D.*?)(\d*)$/
PS: I'm re-factoring Perl code to Java.
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`(?:...)` is non capturing group. – anubhava Oct 11 '14 at 11:25
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`123 fdhdhf234` for this input , 1st capturing group index contains 123 and the second capturing group index contains 234. – Avinash Raj Oct 11 '14 at 11:31
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@AvinashRaj When i run this code `perl -e '$string="123fdhdhf234"; $string =~ m/^(\d*)(?: \D.*?)(\d*)$/; print $1; print $2;'` Nothing gets printed. – Saravanan Oct 11 '14 at 12:04
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Yes, because there is no space after the first three digits. This regex `^(\d*)(?: \D.*?)(\d*)$` would match the string only if it's starts with a number followed by a space or a space. – Avinash Raj Oct 11 '14 at 12:06
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Sorry for the bother.. Thanks @AvinashRaj ! – Saravanan Oct 11 '14 at 12:08
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You should always check if match was suceful `perl -e '$string="123fdhdhf234"; print $1, $2 if $string =~ m/^(\d*)(?: \D.*?)(\d*)$/;` – mpapec Oct 11 '14 at 12:11
1 Answers
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It means that it is not capturing group. After successful match first (\d*) will be captured in $1, and second in $2, and (?: \D.*?) would not be captured at all.
$string =~ m/^(\d*)(?: \D.*?)(\d*)$/
From perldoc perlretut
Non-capturing groupings
A group that is required to bundle a set of alternatives may or may not be useful as a capturing group. If it isn't, it just creates a superfluous addition to the set of available capture group values, inside as well as outside the regexp. Non-capturing groupings, denoted by (?:regexp), still allow the regexp to be treated as a single unit, but don't establish a capturing group at the same time.
