Here's a regex that should get you started: /("[^"]*"|[^\s"]+)/g; This regex makes two following assumptions: a word is a sequence of non-whitespace, non-quote characters, and a quotation cannot contain embedded quotes.

sub split_text {
  $_[0] =~ /("[^"]*"|[^\s"]+)/g;
}

my $text = 'The dog said "Hi, how are you?"  I laughed.';

my @pieces = split_text($text);

print "$_\n" for @pieces;

__END__
The
dog
said
"Hi, how are you?"
I
laughed.
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perfection! muchos gracias!

And here is another way (you can wrap it in a function if you wish) that returns the text inside the quotes without the qoutes:

my $str = qq|"this is one" and "anot""her one" is "" right "her"e|;

my @stuff = map { /([^"]+)"$/ } $str =~ /"([^"]*"|[^\s"]+)/g;

print join("\n", @stuff), "\n";
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The first regular expression returns a list like so:

this is one"
anot"
her one"
"
her"
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and the second regex inside the map removes the left over quote and, as a feature of map, doesn't return the 'empty' ones - so the final list looks like:

this is one
anot
her one
her
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I am sure that this process could be compacted into yet another single regex, but it eludes me.

Jeff

R-R-R--R-R-R--R-R-R--R-R-R--R-R-R--
L-L--L-L--L-L--L-L--L-L--L-L--L-L--

and, as a feature of map, doesn't return the 'empty' ones

er, um. map does return empties; grep does not. however, i think what's going on in that case is that map is returning an empty list, which then gets "lost" in the assignment into a list, because of how arrays flatten in assignment. e.g.,

@a = ("1", "2", "3"), "4", ("5", "6"), (), "7";
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is equivalent to

@a = qw(1 2 3 4 5 6 7)
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.

here's a one-liner that demonstrates that map returns empties:

perl -e '@foos = map { /[13579]/ ? "foo" : "" } (0..50); print join "\
+n", @foos; print "\n\n"'
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and one the demonstrates that grep doesn't ¹:

perl -e '@foos = grep { /[13579]/ ? "foo" : "" } (0..50); print join "
+\n", @foos; print "\n\n"'
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... and that lists get squashed if map happend to be returning lists:

perl -e '@foos = map { /[13579]/ ? ("foo",$_) : () } (0..50); print jo
+in "\n", @foos; print "\n\n"'
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.

[1] well, it isn't as easy as that. grep returns a list of the values on the rhs that make the grep evaluate to true. but if you can use it like a map by assigning into $_, if you really want to:

perl -e '@foos = grep { $_= (/[13579]/ ? "foo" : "") } (0..50); print 
+join "\n", @foos; print "\n\n"'
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but i'm digressing again.
.