Quote:
Originally Posted by
pludi
Maybe the gurus at the PerlMonks - The Monastery Gates can help.
Thanks, pludi, I wrote both an email to Jeffrey Friedl, and a post on perlmonks.
Here's my answer to Perl gurus from perl.org:
===
Let's present the re
'ab' =~ /((\w+)(?{print defined $2 ? "\$2=$2\n" : "\$2 not defined\n"})){2}/;
as
((\w+)(?{print...}))((\w+)(?{print...}))
\w{2} is 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+) captures the letter a:
((\w+) | (?{print...}))((\w+)(?{print...}))
$2 must receive the value a, but in current version of Perl $2 receives
undefined... Why? Probably, two values of undefined are stored in $2 as in a stack,
then last value is removed from the stack, and $2 again equal undefined?
Here eval must print $2=a.
5. Second copy of (\w+) captures the letter b:
((\w+)(?{print...}))((\w+) | (?{print...}))
Eval prints $2=b. Match successfull.
Do you see any mistake in this reasoning?
===
Latest editing:
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.
Sorry for my poor English.
===
After previous post I think again and now I think than intuitively $2=undefined should be incorrect, and $2=a correct.
After that I've received an email from guru of regex Jeffrey Friedl (regex.info):
---
Hi Serge,
I've been thinking about this for a while, and as far as I can tell it does seem
to be a bug. By definition, $2 must be defined before the (?{...}) can run.
It's probably a problem with how it backtracks. I'd suggest filing a bug report..
---
Splitting the regex:
((\w+)(?{print...}))((\w+)(?{print...}))
is wrong, really the regex is not split.
After (\w+) captures all the string:
(\w+)) | {2}
we see, that second repetition of \w not match. We do backtracking and enter second parentheses going from right to left:
(/w | )+
In this case the regex engine (as I think) set $2=undefined, but why? Intuitively it seems set $2=undefined should do after we leave the open second parenthesis going from right to left.
