in reply to
Perl Idioms Explained - @ary = $str =~ m/(stuff)/g
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/?