in reply to Re^9: Understanding a portion of perlretut
in thread Understanding a portion on the Perlretut

In the first case, the engine starts from the left and finds a match, repeating (\w\w\w) 3 times: 
uvXYZdabcXYZfg
^        ^   
|        |
A        B

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The engine than starts to match at B + 1, and finds no such a match.

In the second case, the engine starts from the left as well, but finds no match: 

uvXYZdabcXYZfg
^
|
A

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So, it moves to A + 1 (still no match), and then A + 2, where it can match with (abc)* repeating zero times: 

uvXYZdabcXYZfg
  ^
  |
 A=B

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After matching, it continues (because of /g) to B + 3 (no match), and at B + 4 it finally succeeds with 

uvXYZdabcXYZfg
      ^  ^
      |  |
      A  B

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Better now? 

($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,

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Re^11: Understanding a portion of perlretut
by Athanasius (Archbishop) on Dec 10, 2015 at 13:27 UTC

    Yes, thanks, I’m starting to see some light. :-) But:

    In the first case, the engine starts from the left and finds a match, repeating (\w\w\w) 3 times:

    Makes sense, but then, if the $1 match effectively ends up as (\w\w\w){3}, shouldn’t $1 contain uvXYZdabc? Or, conversely, if the ? quantifier following the * causes it to find the minimum string in uvXYZdabc matching (\w\w\w)*, shouldn’t that be "", the null string? Why does $1 match abc here?

    Athanasius <°(((>< contra mundum Iustus alius egestas vitae, eros Piratica,

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      The repeated capturing overwrites the value in $1. If you want the repetition inside of a capture variable, you'll have to use something like ((?:\w\w\w)*?).

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