$2=ab
$2=a
$2=b
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Please expound on why you believe this should be the output. To my understanding, the regular expression specification defines output but not method. If there is a specification or archived developer discussion I am unaware of, I would appreciate the citation. Otherwise, I do not see a compelling argument for "it should be the last thing matched by those 'physical' parentheses" over my proposal.

A successful capture should not be undef. Whether it's documented or not, $1, etc are intended to be available to (?{}) and (??{}) blocks, so they must be set ASAP.

1. \w+ matches "ab"
2. () sets $2
3. Print $2
4. Second at \w+ fails to match: backtrack
5. \w+ matches "a"
6. () sets $2 (This isn't happening)
7. Print $2
8. \w+ matches "b"
9. () sets $2
10. Print $2

I do not see a compelling argument for "it should be the last thing matched by those 'physical' parentheses" over my proposal.

If that's true, then you would expect either of

$2=ab  # Only thing matched
$2=a   # Only thing matched
$2=a   # First thing matched
[download]

and

$2=ab  # Only thing matched
$2=a   # Only thing matched
$2=b   # Last thing matched
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and the code is still buggy as it produces neither.

Clearer bug demonstration:

$ perl -e'"ab" =~ /((\w+)(?{print defined $^N ? "\$^N=$^N\n" : "\$^N n
+ot defined\n"})){2}/;'
$^N=ab
$^N not defined
$^N=b
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It seems, I've mistaken. Here's my correction to my previous reasoning.

Let's present the re

'ab' =~ /((\w+)(?{print defined $2 ? "\$2=$2\n" : "\$2 not defined\n"})){2}/;

as

((\w+)(?{print...}))((\w+)(?{print...}))

Is \w{2} equivalent to \w\w, right? But we assume that the second copy of the re produces also the same $1 and $2 (not $3 and $4). Current position in the re marked with |.

1. First (\w+) captures all the text:
((\w+) | (?{print...}))((\w+)(?{print...}))
$2 receives the value 'ab', eval prints $2=ab.

2. Then we enter second copy of (\w+):
((\w+)(?{print...}))(( | \w+)(?{print...}))
$2 (and also $+, $^N, \2) receives the value undefined.

3. We see that \w not match. We do backtracking:
((\w+ | )(?{print...}))((\w+)(?{print...}))
We enter first copy of (\w+) from right to left, and $2 again receives the value undefined.

4. \w+ gives back the letter b (but $2 remains undefined, because we did not come left of the opening parenthesis for $2):
(( | \w+(?{print...}))((\w+)(?{print...}))
$2 remains undefined.

4. (\w+) captures none, because we did not come left of the opening parenthesis for $2:
((\w+) | (?{print...}))((\w+)(?{print...}))
$2 remains undefined. Eval prints $2=undefined.

5. Second copy of (\w+) captures the letter b:
((\w+)(?{print...}))((\w+) | (?{print...}))
Eval prints $2=b. Match successfull.