Greediness is often misunderstood, but actually that is beside the point considering the regex supplied. After all, what are you saying when you say s/\s+|\n/ /g? Consider...

$data = "\n\n\n";
$data =~ s/\s+/foobar/g;
print "data=>$data\n";
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This outputs "foobar", because \n is whitespace covered by \s.

Both * and + are greedy operators when not flipped to non-greedy by the application of ?. True, the implications of greediness are very important when greedy operators are combined with alternation (as above), but those implications do not come into play given the example. In the given regex, if you used ? to change your greedy + operator to non-greedy, then the RHS of the alternation operation would be applied. Regardless, the effect would be the same (' ' would be substituted for \n versus for \s, but you'd see the same result). Update: Let me strike that last bit. Got effect of ? on alternation messed up. Rest of explanation is fine (I think).

Forgive me if I am missing something (I have the feeling I must be).

On another note (to the original author of the SOPW node), please note that you don't need multi-line matching here. The m and s regex modifiers only have to do with "^", "$", and ".", not with \s.

Yeah, I didn't realise that \s was a whitespace (as opposed to a plain old space) matcher, hence the use of \s+\n.

Maybe you already know that if you want to match a whitespace character except '\n', you could use the [^\S\n] idiom. Where \S matches any non-white space character, \n matches the return, and then take the complement of the set gives you \s but not \n. Handy trick in situations when you want to differentiate between whitespace and \n.



my $text = 'hello
   this is a multiline
   message';

$text =~ s/([^\S\n]+|\n)/$1 eq "\n" ?
                         print "New line\n" :
                         print "Whitespaces\n"; $1/gemi;
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And you get -


New line
Whitespaces
Whitespaces
Whitespaces
Whitespaces
New line
Whitespaces
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