regexp: non-capturing grouping in replacement possible?

diweooy has asked for the wisdom of the Perl Monks concerning the following question:

hi, I am trying to use non-capturing grouping (?:) in an replacement:

my $str = 'hello kitty 003x009 spanish';
$str =~ s/^(?:hello kitty )(\d+)(?:x\d+.*)$/sprintf "%d", $1/e;

print $str;

# result: 3
# expected: 'hello kitty 3x009 spanish'
[download]

I tought that non-capturing grouping maybe does not work with evaluation modifier 'e' but that was not problem:

$str =~ s/^(?:hello kitty )(\d+)(?:x\d+.*)$/XX/;
[download]

produces 'XX' as result. does non-capturing work when replacing?

Comment on regexp: non-capturing grouping in replacement possible? Select or Download Code

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Re: regexp: non-capturing grouping in replacement possible? by moritz (Cardinal) on Nov 26, 2009 at 12:50 UTC
You need look-around assertions, not non-capturing groups. See perlre, search for positive look-ahead assertion and zero-width positive look-behind assertion. A substitution substitutions not only the captured part of the string, but the matched part of the string. Using zero-width assertion you can reduce the matched part of the string to the captured part.	[reply]
Re^2: regexp: non-capturing grouping in replacement possible? by diweooy (Novice) on Nov 26, 2009 at 13:11 UTC
I forgot to mention that I tried it but either I am doing something wrong or ...: `$str =~ s/(?=hello kitty )(\d+)/sprintf "%d", $1/e;` [download] does not do anything. it is probably me but I don't see it.	[reply] [d/l]
Re^3: regexp: non-capturing grouping in replacement possible? by happy.barney (Friar) on Nov 26, 2009 at 13:27 UTC
?= is look ahead. try ?<= (look behind)	[reply]
Re^4: regexp: non-capturing grouping in replacement possible? by moritz (Cardinal) on Nov 26, 2009 at 14:23 UTC
Re^4: regexp: non-capturing grouping in replacement possible? by diweooy (Novice) on Nov 26, 2009 at 14:26 UTC
Re: regexp: non-capturing grouping in replacement possible? by JavaFan (Canon) on Nov 26, 2009 at 13:31 UTC
I'd write that as (untested): `s/^hello kitty \K0([1-9][0-9])(?=x[0-9])/$1/;` [download] It seem that all you want to do is remove leading 0s.	[reply] [d/l]
Re: regexp: non-capturing grouping in replacement possible? by biohisham (Priest) on Nov 26, 2009 at 14:13 UTC
Can not a simpler approach give you what you're looking for? capture the 3 which is preceded by 0s, remove those 0s and preserve the 3.. `my $str = 'hello kitty 003x009 spanish'; $str =~ s/0*(3)/$1/; print $str;` [download] Excellence is an Endeavor of Persistence. Chance Favors a Prepared Mind.	[reply] [d/l] [select]
Re^2: regexp: non-capturing grouping in replacement possible? by diweooy (Novice) on Nov 26, 2009 at 14:21 UTC
thanks but it is not about zeros but about learning. I could remove zeroes with a simple 0. `$str =~ s/^(hello kitty )0(\d+)(x\d+.*)$/$1$2$3/;` [download]	[reply] [d/l]
Re^3: regexp: non-capturing grouping in replacement possible? by happy.barney (Friar) on Nov 26, 2009 at 16:14 UTC
learning? this one also removes any leading zero of number after "hello kitty" :-) `s/(?<=hello kitty )0*(?=\d(?!\d))//` [download] Update: oops, small mistake, `\b` replaced with `(?!\d)`	[reply] [d/l] [select]