Errm, jeffa, that woud screw the my ($start,$middle,$end) part up.
Suggested alternatives:

# first one
my ($start, $middle, undef, $end) = $string =~ /^(.*?)(($pattern)+)(.*
+?)$/g;

# second one, with non-capturing parens -- I like it better
my ($start,$middle,$end) = $string =~ /^(.*?)((?:$pattern)+)(.*?)$/g;
[download]

gri6507, note that /$pattern+/ is exactly the same as /AB+/ (well, at least for $pattern = 'AB'), which is, by definition /A(:?B)+/.
Hope this helped.
CombatSquirrel.

Update: Arghh - wrong order for capturing and non-capturing parens. Fixed.

Update 2: jeffa is right. The following RegEx should do the trick:

$pattern = 'AB';
'BABABABBB' =~
/
   ^                   # start at beginning of line
   (                   # capture to $1
      .*?                 # a number of character, but as few as possi
+ble ...
      (?<!$pattern)       # ... which may not contain $pattern
   )
   (                   # capture to $2
      (?:$pattern)+       # multiple occurences of $pattern
         |                   # OR
      (?!.*?$pattern)     # nothing, BUT there may be no $pattern in t
+he rest
                          #    of the string
   )
   (.*)                # capture rest to $3
/x ;
print "$1<$2>$3$/";
__END__
prints "B<ABABAB>BB"
[download]

I'm open for any suggestions, and yes, I do know Mastering Regular Expressions, I just forgot half (the important half) of it.

Update 3 (Explanation): The RegEx engine tries to match at the earliest possible position. Therefore it will always match nothing to be captured in $1 (non-greedy dot-star), the highest possible number of following pattern matches (greedy star) and then the rest. Meaning, if the first pattern does not begin at the first character, $2 will also be empty (after all a star does not have to match) and the rest is slurped into $3. Bon appetit!