If you make the last parenthesized match non-greedy by removing the trailing question mark, you can leave out the final $.

I recommend using a line like the following to show what you've captured, just in case you have whitespace: print "1: ->$1<-\n2: ->$2<-\n3: ->$3<-\n";

C:\>perl -We "$_='blah, blah2';$OPnot='-';$OPor=',';$OPand='\+'; 
/^(.*)\s*?($OPnot|$OPor|$OPand)\s*?(.*?)$/; print qq[1='$1', 2='$2', 3
+='$3']"
1='blah', 2=',', 3=' blah2'
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BTW: If you want \s*? to match anything, you should remove the question mark. *? will happily match the gap between chars.
(* matches zero or more, but ? tells it to match as little as possible, aka zero.)

Adam, that's not quite accurate about the '?'. If a question mark follows a quantifier (*?, +?, {min, max}? or ??) in a regex, it makes it "non-greedy". Consider the following code.

# 3 spaces, a tab, 3 more spaces, another tab and 3 more spaces (repre
+sent by chr() for clarity)
$test = chr(32)x3 . chr(9) . chr(32)x3 . chr(9) . chr(32)x3;

($first = $1, $second = $2) if $test =~ /(\s*)\t(\s*)/;
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In this case, the first (\s*) will be greedy and attempt to match as many characters as possible. $first will contain 3 spaces, a tab, and 3 more spaces. $second will contain 3 spaces. However, by adding the question mark, we make it non-greedy.

($first = $1, $second = $2) if $test =~ /(\s*?)\t(\s*)/;
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This means that (\s*?) attempt the smallest match possible that satisfies that above regex. In this case, $first contains 3 spaces and $second contains 3 spaces, a tab, and 3 more spaces. The '?' does not mean "aka zero".

Incidentally, most regexes ending in (.*?)$/ (like the one in the original post) have a superfluous ? because there is no way to make that statement non-greedy, since it's forced to match to the end.

You are correct, perhaps I should have been more clear. The regex that we were discussing ends with \s*?(.*?)$/; which is somewhat different from your example. Here it is matching the fewest spaces followed by the fewest 'anything but newlines' to the end of the string. Since the . will match white space, the \s*? will match nothing. Always. But thank you for your clarification of the more generic case.

\s*? is set that way on purpose in case the words have no whitespace between them.

The question mark in \s*? is not necessary if you are doing that "in case the words have no whitespace between them." The * quantifier matches zero or more of whatever it is quantifying.

$test = "az";
print "Good\n" if $test =~ /a\s*z/;
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The above regex sees an 'a', followed by zero spaces, followed by a 'z'. Since this matches the value of $test, it prints "Good\n".

Cheers!

$OPand was set to '\s+', as this was how the input string is set. This is also why I was using \s*?, believing it would capture any whitespace in the case of the $OPor or $OPnot separators. However, perl was evaluating "blah, blah2" and returning $1 = 'blah,', $2 = ' ', and $3 = 'blah2'. The reason I erroneously assumed the regexp was destroying 'blah2' is the script loops, evaluating $1 as the test string. The handler for ($2 =~ /^\s+$/) was written incorrectly, so the script just skipped over the first iteration. It worked on the second iteration, however, evaluating "blah,", and thus finding $1 = "blah", $2 = ",", and $3 = "".

Again, thanks for the pointers--they definitely helped me figure out what I'd done incorrectly.