From 9201e22a70ab4dc37c2d5996d18fa09ef00ea706 Mon Sep 17 00:00:00 2001 From: jim Date: Sat, 21 Mar 2015 10:45:35 -0400 Subject: [PATCH] Explain why blackhole and blackhole () can have the same type --- exercises/assignment5_answers.mdwn | 2 ++ 1 file changed, 2 insertions(+) diff --git a/exercises/assignment5_answers.mdwn b/exercises/assignment5_answers.mdwn index 5425ac51..f62f8b4f 100644 --- a/exercises/assignment5_answers.mdwn +++ b/exercises/assignment5_answers.mdwn @@ -778,3 +778,5 @@ and that `bool` is any boolean expression. Then we can try the following: match b with true -> y () | false -> n () that would arguably still be relying on the special evaluation order properties of OCaml's native `match`. You'd be assuming that `n ()` wouldn't be evaluated in the computation that ends up selecting the other branch. Your assumption would be correct, but to avoid making that assumption, you should instead first select the `y` or `n` result, _and then afterwards_ force the result. That's what we do in the above answer. + + Question: don't the rhs of all the match clauses in `match b with true -> y | false -> n` have to have the same type? How can they, when one of them is `blackhole` and the other is `blackhole ()`? The answer has two parts. First is that an expression is allowed to have different types when it occurs several times on the same line: consider `let id x = x in (id 5, id true)`, which evaluates just fine. The second is that `blackhole` will get the type `'a -> 'b`, that is, it can have any functional type at all. So the principle types of `y` and `n` end up being `y : unit -> 'a -> 'b` and `n : unit -> 'c`. (The consequent of `n` isn't constrained to use the same type variable as the consequent of `y`.) OCaml can legitimately infer these to be the same type by unifying the types `'c` and `'a -> 'b`; that is, it instantiates `'c` to the functional type had by `y ()`. -- 2.11.0