In reply to the update:
// does not match "anything". It is overloaded to reapply the last successfully matched pattern, in your case, /b/. You wouldn't notice that because all your test data past the successful match has a "b" in it… Changing that clearly reveals faulty logic: once the right side is true, and flipflop starts over on the left side again.
$ perl -le'for( qw/foo bar baz blah/ ){ print int scalar( /b/ .. // )
+}'
0
1
1
1
$ perl -le'for( qw/foo bar baz quux/ ){ print int scalar( /b/ .. // )
+}'
0
1
1
0
Now let's take a short trip to perlop:
The value returned is either the empty string for false, or a sequence number (beginning with 1) for true. The sequence number is reset for each range encountered. The final sequence number in a range has the string "E0" appended to it, which doesn't affect its numeric value, but gives you something to search for if you want to exclude the endpoint. You can exclude the beginning point by waiting for the sequence number to be greater than 1.
You want a right side that does not succeed. Because once it succeeds, you start over on the left side again!
Now a simple, correct match-always pattern is /^/ or /(?=)/, and a simple match-never pattern is /$^/ or /(?!)/. Indeed, using a match-never pattern:
$ perl -le'for( qw( foo bar baz quux ), "" ){ print int scalar( /b/ ..
+ /$^/ ) }'
0
1
2
3
4
$ perl -le'for( qw( foo bar baz quux ), "" ){ print int scalar( /baz/
+.. /$^/ ) }'
0
0
1
2
3
Well, that sequence looks right. So that means you can say:
log_failure()
unless grep {
( exists $hash{$_} and ( do_something($_), 1 ) )
.. /$^/
} @options;
Makeshifts last the longest.
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