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Hi, I have a question on the capturing parentheses. I have made the following test sample to explore the necessity of the parentheses:

$str = "Ab stuff Cd stuff Ef stuff"; # case 1 @ary1 = $str =~ m/(stuff)/g ; # case 2 @ary2 = $str =~ m/(?:stuff)/g ; # case 3 @ary3 = $str =~ m/stuff/g ; print "\@ary1 = @ary1\n"; print "\@ary2 = @ary2\n"; print "\@ary3 = @ary3\n";
All three cases return the same result. Explanation for case 1 is covered in earlier posts. However I am puzzled by the use of parentheses in the example, so I added ?: to it to tell the regular expression to forget the value in the capture parentheses if any. The result is the same! So the regular expression is not acting on the $1 variable captured by the parentheses at all. So I eliminated the parentheses totally, I still get the same result.

Ok, my instinct tells me that this Perl idiom is acting on the behaviour of m//g, or more specific the g modifier. It seems the g modifier introduces it's own pattern matching memory behaviour and discards the regular expression memory in some cases.

I looked up the perldoc, which states:
The /g modifier specifies global pattern matching--that is, matching as many times as possible within the string. How it behaves depends on the context. In list context, it returns a list of the substrings matched by any capturing parentheses in the regular expression. If there are no parentheses, it returns a list of all the matched strings, as if there were parentheses around the whole pattern.

Ok, my question is, what is the expected behaviour of the g modifier? Why is the /g modifier capturing the value that I want it to forget (with ?:)? Is it a feature or bug? Or perhaps /(?:pattern)/ is equivalent to /pattern/?

In reply to Re: Perl Idioms Explained - @ary = $str =~ m/(stuff)/g by Roger
in thread Perl Idioms Explained - @ary = $str =~ m/(stuff)/g by tachyon

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