That isn't how it works. HVM threads work transparently, you don't need to think about them. The same data structure *can* be visible to multiple threads under

That isn't how it works. HVM threads work transparently, you don't need to think about them. The same data structure *can* be visible to multiple threads under some conditions, through cloning; for example, consider the following program: (Sum Nil ) = 0 (Sum (Cons x xs)) = (+ x (Sum xs)) (Head (Cons x xs)) = x (Main x) = let xs = (Cons 1 (Cons 2 (Cons 3 Nil))) (Pair (Sum xs) (Head xs)) The result of this program is `(Pair 6 1)`. Since `Pair` is the topmost (root) node of the normal form, each of its elements will be computed in a separate thread. Because of that, the same `[1,2,3]` list will be accessed by each thread. But that's not actually a shared reference, because values only exist in one place; instead, the lazy cloner makes copies of the list, layer-by-layer, and sends these copies to the thread that will handle them. Note that if the same expression was written in a deeper position of the normal form, it would be evaluated sequentially. This is what avoids the fine-grained parallelism issue. It also means that in many cases HVM will not be as parallel as it could, though! But when the parallelism kicks in, it is always productive.