You are putting the result of the operation 7,2,3 which is 7.

No, it is not. Consider:

perl -le 'sub foo { 7,2,3 } $foo = foo(); print $foo'
3
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It's a precedence problem as other posters pointed out.

update Well, actually... the above example is misleading, since the comma operator is used in list context here - subroutines always return lists (if not declared with a single ($) as prototype, that is). And evaluating a list in scalar context gives its last element - which is just what the comma operator does ;-) A better example might be

print not 1,0;
print $/;
__END__
1
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The not operator gets the 0 and turns it to 1.

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}