Re^2: Greedy modifier found to be working non-greedy in a named group

[rkabhi]

@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?

[Corion]

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.

[hippo]

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.

[LanX]

> 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>

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

°) yes there is an maximal upper bound in range quantifiers, which I expected to be around 2**16, so it's an incomplete analogy

[rsFalse]

> ...and that e* means e{0}|e+ ( or something like e{0,32766} ° ).

> So you are actually matching e{0} before all!

And if the sentence begins with "e.."? :)

[LanX]

Good catch!

The order must be reversed to reflect the greed.

  DB<8> x 'eeeA1234eB' =~ /(e+|e{0})(.)/
0  'eee'
1  'A'
  DB<9> x 'A1234eB' =~ /(e+|e{0})(.)/
0  ''
1  'A'
  DB<10> x 'A1234eB' =~ /(e+)(.)/
0  'e'
1  'B'
  DB<11>

Thanks!

Updates

Improved demo code

[rkabhi]

That was very good explanation.

Thanks a lot !!

Regards,
Abhishek

[rkabhi]

Thanks, its clear now.