... in this case explicit \n matches where dot with s mod doesn't.

That's because  \n$ requires a match with newline, but  .$ allows the leftmost position of a match with dot (with /s) to be before the newline. Dot will match newline in the presence of  $ if it is the only match available: 

c:\@Work\Perl\monks>perl -wMstrict -le
"use Data::Dump qw(pp);
 ;;
 for my $s (qq{yz}, qq{yz\n}, qq{\n}) {
   $s =~ m{ (.) $ }xms;
   printf qq{in %s matched %s \n}, pp($s), pp($1);
   }
"
in "yz" matched "z"
in "yz\n" matched "z"
in "\n" matched "\n"

[download]
(The  /m modifier makes no difference in these example strings.)

The thing to remember about regular expressions is that there are a lot of things to remember about regular expressions. If you have a chance to reduce the amount of stuff to remember, even if only by a little, take it. That's why I advise (per TheDamian's regex PBPs) using  \A \z \Z for all your start- and end-of-string anchoring needs, and using  ^ $ only for embedded newline matching.

... inconsistency ...

For me, it's not so much inconsistency as mind-boggling complexity. And again, I come back to the point that if you can reduce the complexity of what you're dealing with even a little, you're ahead of the game.

Give a man a fish:  <%-{-{-{-<

In reply to Re^3: How to match last character of string, even if it happens to be a newline? by AnomalousMonk
in thread How to match last character of string, even if it happens to be a newline? by Allasso

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