Re^2: list of four digit lock combinations without repeated digits

I did ask for interesting solutions. It's going to take me a while to understand how this works. I found in perlvar a description of @- as being an array of the indexes of the matches. Thanks for posting.

@-
$-[0] is the offset of the start of the last successful match. $-[n] i
+s the offset of the start of the substring matched by n-th subpattern
+, or undef if the subpattern did not match.

Thus, after a match against $_ , $& coincides with substr $_, $-[0], $
++[0] - $-[0] . Similarly, $n coincides with substr $_, $-[n], $+[n] -
+ $-[n] if $-[n] is defined, and $+ coincides with substr $_, $-[$#-],
+ $+[$#-] - $-[$#-] . One can use $#- to find the last matched subgrou
+p in the last successful match. Contrast with $#+ , the number of sub
+groups in the regular expression. Compare with @+ .

This array holds the offsets of the beginnings of the last successful 
+submatches in the currently active dynamic scope. $-[0] is the offset
+ into the string of the beginning of the entire match. The nth elemen
+t of this array holds the offset of the nth submatch, so $-[1] is the
+ offset where $1 begins, $-[2] the offset where $2 begins, and so on.
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Re^3: list of four digit lock combinations without repeated digits by tybalt89 (Monsignor) on Jun 22, 2018 at 22:45 UTC
Here's another version without the @- Just match any four of the digits (from an ordered string). `#!/usr/bin/perl # https://perlmonks.org/?node_id=1217042 use strict; use warnings; '0123456789' =~ /(.).?(.).?(.).?(.)(?{print "$1$2$3$4\n"})(FAIL)/;` [download]	[reply] [d/l]