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!

In reply to RE: RE: Re: Regexp oddity by Ovid
in thread Regexp oddity by daemon23

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