... match the longest "or" possible ...

The Perl regex engine matches the leftmost longest substring for a given pattern; leftmost trumps longest. If you want to match the longest substring that occurs anywhere in the string (which, IIRC, is the POSIX definition of matching), you have to jump through a hoop or two: 

>perl -wMstrict -le
"my $s = 'x s m e f s f pl s f x';
 ;;
 $s =~ qr/^.+(s m e f|s f pl|s f).+$/;
 print qq{'$1'};
 print qq{ -------};
 ;;
 my $alt = qr{ s[ ]m[ ]e[ ]f | s[ ]f[ ]pl | s[ ]x }xms;
 for my $s (
   'x  s m e f  s f pl  s f  x',  'x  s f  s f pl  s m e f  x',
   'x  s f  s m e f  s f pl  x',
   ) {
   my ($longest) =
     reverse sort { length($a) <=> length($b) }
     $s =~ m{ (?<= .) $alt (?= .) }xmsg
     ;
   print qq{'$longest'};
   }
"
's f'
 -------
's m e f'
's m e f'
's m e f'

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I think there is another, possibly more succinct, solution using the 5.10+ Special Backtracking Control Verbs, but I can't remember it ATM.

Update: Well, this doesn't use backtracking control verbs quite the way I imagined, but at least it gets rid of the list-making and sorting: 

>perl -wMstrict -le
"my $alt = qr{ s[ ]m[ ]e[ ]f | s[ ]f[ ]pl | s[ ]f }xms;
 my $rex = qr{ (?<= .) $alt (?= .)                }xms;
 ;;
 for my $s (
   'x  s m e f  s f pl  s f  x',  'x  s f  s f pl  s m e f  x',
   'x  s f  s m e f  s f pl  x',  'x  s f  s f  x',
   'x s s x',  's f',  's m e f',
   ) {
   local our $longest;
   local our $offset;
   use re 'eval';
   $s =~ m{
     ($rex)
     (?{ ($longest, $offset) = ($1, $-[1])
       if ! defined($longest) || length($1) > length($longest)
       })
     (*SKIP)(*FAIL)
     }xmsg;
   print qq{'$s' -> '$longest' at $offset} if defined $longest;
   }
"
'x  s m e f  s f pl  s f  x' -> 's m e f' at 3
'x  s f  s f pl  s m e f  x' -> 's m e f' at 16
'x  s f  s m e f  s f pl  x' -> 's m e f' at 8
'x  s f  s f  x' -> 's f' at 3

[download]

Note that in the string  'x  s f  s f  x' the left-most occurrence of 's f' is found. This can be changed to the right-most by changing the comparison operator in  length($1) > length($longest) to >= instead.

In reply to Re: match longest sequence in an "or" RE by AnomalousMonk
in thread match longest sequence in an "or" RE by silentius

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