To get at your question more directly, it is true that closures in Perl do not have to be anonymous, and as others have said, yes technically your example is a closure. But the canonical case for named closures is something like this in Lisp (forgive my somewhat rusty Lisp):

(defun foo (n) 
   (labels ((f (i) (+ i m)))
           f)
)
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and you might be tempted to write the Perl equivalent like this:

sub foo {
   my $n = shift;
   sub f { 
       my $i = shift;
       return $i+$n;
   }
   return \&f;
}
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But this will not work, because the inner sub "f" is really global and the closure won't behave like you want it to. The solution is to make f anonymous:

sub foo {
   my $n = shift;
   my $f = sub { 
       my $i = shift;
       return $i+$n;
   };
   return $f;
}
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And this is the main reason why closures in Perl are usually anonymous.