2. because thats how a variable is compared in regex

No, the variable $constant will be interpolated into the regex without the parentheses (round brackets). Consider:

1:31 >perl -wE "my $constant = 'fred'; my $string = 'fredflintstone'; + print qq[found $1\n] if $string =~ /^$constant/;" Use of uninitialized value $1 in concatenation (.) or string at -e lin +e 1. found 1:32 >perl -wE "my $constant = 'fred'; my $string = 'fredflintstone'; + print qq[found $1\n] if $string =~ /^($constant)/;" found fred 1:32 >

If the regex matches, any parentheses capture their contents into the special variables $1, $2, etc. But this incurs a performance penalty:

WARNING: Once Perl sees that you need one of $&, $`, or $' anywhere in the program, it has to provide them for every pattern match. This may substantially slow your program. Perl uses the same mechanism to produce $1, $2, etc, so you also pay a price for each pattern that contains capturing parentheses. (“Capture groups” in perlre#Regular-Expressions)

Since you’re not using $1, the capturing parentheses aren’t needed.

Hope that helps,

Athanasius <°(((><contra mundum Iustus alius egestas vitae, eros Piratica,


In reply to Re^3: Issue with regex matching by Athanasius
in thread Issue with regex matching by sowais

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