Greedy modifier found to be working non-greedy in a named group

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

Dear Monks,

I am new to named groups in Perl regex. I am using perl 5.16.0

Below code works as expected:

$_ = "This is a teeeext for testting";
/(?<char>.*)/
     and print "'$+{char}' is matched pattern\n";
# This prints 'This is a teeeext for testting' is matched pattern
[download]

Replacing '.' with 'e' in regular expression above leaves matched pattern as blank at the beginning:

$_ = "This is a teeeext for testting";
/(?<char>e*)/
     and print "'$+{char}' is matched pattern\n";
# This prints '' is matched pattern
[download]

I was expecting output of above code to be => 'eeee' is matched pattern Please enlighten why the output is different? Is the * modifier becoming non-greedy or is there any other way to understand this?

Comment on Greedy modifier found to be working non-greedy in a named group Select or Download Code

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Re: Greedy modifier found to be working non-greedy in a named group by tybalt89 (Monsignor) on Nov 29, 2019 at 10:18 UTC
`$_ = "This is a teeeext for testting"; /(?<char>e)/ and print "'$+{ch +ar}' is matched pattern at $-[0]\n";` [download] Outputs: `'' is matched pattern at 0` [download] e matches at the beginning of the string, try e+	[reply] [d/l] [select]
Re^2: Greedy modifier found to be working non-greedy in a named group by rkabhi (Acolyte) on Nov 29, 2019 at 11:44 UTC
@tybalt89 I have tried e+ already and I know that it works. But e* should also have worked because '' is a greedy modifier unless it is succeeded by a '?'. So, in my example, both e and e+ should have given same output. Why is the difference in output?	[reply]
Re^3: Greedy modifier found to be working non-greedy in a named group by Corion (Patriarch) on Nov 29, 2019 at 12:13 UTC
The greedy operators still implement a leftmost-longest strategy. That means that the leftmost match will win even if you find a longer match later. `e*` can match zero e. The leftmost place where you can match zero e is at the start of the string.	[reply] [d/l]
Re^3: Greedy modifier found to be working non-greedy in a named group by hippo (Bishop) on Nov 29, 2019 at 12:13 UTC
A greedy match is the longest representation of the first match. In the case of `/e*/` the first match is the start of the string, and is zero characters long because the first character is not an 'e'. In the case of `/e+/` the first match starts at the first 'e' and continues until it finds a non-'e' character.	[reply] [d/l] [select]
Re^3: Greedy modifier found to be working non-greedy in a named group by LanX (Saint) on Nov 29, 2019 at 12:51 UTC
> both e and e+ should have given same output. Why is the difference in output?* Good question, I think many are confused. It's important to understand that empty matches exist and that `e` means `e{0}\|e+` ( or something like `e{0,32766}` ° ). So you are actually matching `e{0}` before all! Printing out the match position of a capture group via `@+` helps demonstrating it DB<1> $_ = "This is a teeeext for testting"; DB<2> ;/(?<char>e)/ and print "'$+{char}' is matched pattern at p +os $+[0]\n"; '' is matched pattern at pos 0 DB<3> ;/(?<char>e{0})/ and print "'$+{char}' is matched pattern at + pos $+[0]\n"; '' is matched pattern at pos 0 DB<4> ;/(?<char>e+)/ and print "'$+{char}' is matched pattern at p +os $+[0]\n"; 'eeee' is matched pattern at pos 15 DB<5> ;/(?<char>e{1,32766})/ and print "'$+{char}' is matched patt +ern at pos $+[0]\n"; 'eeee' is matched pattern at pos 15 DB<6> [download] update Actually `$+[0]` gives you the end of the first match, `$-[0]` will give you the start. `e` is not limited in my Perl version but the upper bound in `e{,}` is. couldn't find positions of named groups Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery FootballPerl is like chess, only without the dice} °) yes there is an maximal upper bound in range quantifiers, which I expected to be around 2*16, so it's an incomplete analogy	[reply] [d/l] [select]
Re^4: Greedy modifier found to be working non-greedy in a named group by rsFalse (Chaplain) on Nov 30, 2019 at 15:05 UTC
Re^5: Greedy modifier found to be working non-greedy in a named group by LanX (Saint) on Nov 30, 2019 at 20:19 UTC
Re^4: Greedy modifier found to be working non-greedy in a named group by rkabhi (Acolyte) on Nov 29, 2019 at 12:59 UTC
Re^3: Greedy modifier found to be working non-greedy in a named group by rkabhi (Acolyte) on Nov 29, 2019 at 12:43 UTC
Thanks, its clear now.	[reply]
Re: Greedy modifier found to be working non-greedy in a named group by rkabhi (Acolyte) on Nov 29, 2019 at 12:52 UTC
To conclude, the explanation of leftmost first and then longest will be valid even if I didn't use named groups. So, for example, `$_ = "This is a teeeext for testting"; /(e*)/ and print "'$1' is matched pattern\n"; # gives '' is matched pattern` [download] and `$_ = "This is a teeeext for testting"; /(e+)/ and print "'$1' is matched pattern\n"; # gives 'eeee' is matched pattern` [download] So, this behavior is generic.	[reply] [d/l] [select]
Re^2: Greedy modifier found to be working non-greedy in a named group by haukex (Archbishop) on Nov 29, 2019 at 21:08 UTC
the explanation of leftmost first and then longest will be valid even if I didn't use named groups. So, this behavior is generic. Yes, perlre says: In Perl the groups are numbered sequentially regardless of being named or not. Thus in the pattern `/(x)(?<foo>y)(z)/` [download] `$+{foo}` will be the same as `$2`, and `$3` will contain `'z'` And the description of `(?\|pattern)` says: Named captures are implemented as being aliases to numbered groups holding the captures Also, I remember finding the description of the left-to-right operation of the regex engine in the Camel quite enlightening.	[reply] [d/l] [select]


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Greedy modifier found to be working non-greedy in a named group

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