> I think I've understood now 👍

maybe this helps, you can reproduce it in the debugger started with perl -de0

  DB<22> p $_ = join "," , A..E
A,B,C,D,E
  DB<23> p m/(,.*,)/  # longest possibility from first comma to last c
+omma
,B,C,D,
  DB<24> p m/(,.*?,)/ # shortest possibility from first comma to next 
+comma
,B,
  DB<25> p m/(,.*$)/  # longest possibility from first comma to end of
+ line
,B,C,D,E
  DB<26> p m/(,.*?$)/ # shortest possibility from first comma to end o
+f line
,B,C,D,E
  DB<27>
[download]

the regex-engine tries to find a solution step by step:

• first it tries the first pattern, here "," = comma
• then it matches "." = all as many times like quantified ( "*" or "*?" )
• till it matches the next pattern ( "," or "$" = EOL )
• IFF not all criteria can be met, it'll try to start anew from the next comma, and so on

The problem with your regex was, that it was already matching from the leftmost comma.

But all solutions provided by other monks made sure that only the rightmost comma allowed to be a match.

For instance

  DB<27> p m/(,[^,]*)$/ # comma followed by non-commas till EOL
,E
[download]

The engine will actually try to first match all other commas to the left but always fail because it encounters other commas before reaching the EOL.

we can actually make the regex display it's intermediate attempts to match while "backtracking"

  DB<32> ; m/(,[^,]*) (?{say $1}) $/x #show all intermediate attempts 
+to match $1 until it doesn't fail
,B
,
,C
,
,D
,
,E
  DB<33>
[download]

The difference with non-greedy quantifier *? matching is that the engine goes from shortest to longest attempts while backtracking

  DB<34> ; m/(,[^,]*?) (?{say $1}) $/x #show all intermediate attempts
+ to match $1
,
,B
,
,C
,
,D
,
,E

  DB<35>
[download]

Is this clearer now? :)

HTH!

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