Question

OCaml gives function `A -> 1 | _ -> 0 the type [> `A] -> int, but why isn’t that [> ] -> int?

This is my reasoning:

• function `B -> 0 has type [<`B] -> int. Adding a `A -> 0 branch to make it function `A -> 1 | `B -> 0 loosens that to [<`A|`B] -> int. The function becomes more permissive in the type of argument it can accept. This makes sense.
• function _ -> 0 has type 'a -> int. This type is unifiable with [> ] -> int, and [> ] is an already open type (very permissive). Adding the `A -> 0 branch to make it function `A -> 1 | _ -> 0 restricts the type to [>`A] -> int. That doesn’t make sense to me. Indeed, adding still another branch `C -> 1 will make it [>`A|`C] -> int, further restricting the type. Why?

Note: I am not looking for workarounds, I’d just like to to know the logic behind this behavior.

On a related note, function `A -> `A | x -> x has type ([>`A] as 'a) -> 'a, and while that is also a restrictive open type for the parameter, I can understand the reason. The type should unify with 'a -> 'a, [>` ] -> 'b, 'c -> [>`A]; the only way to do it seems to be ([>`A] as 'a) -> 'a.

Does it exist a similar reason for my first example?

Answer 1

A possible answer is that the type [> ] -> int would allow an argument (`A 3) but this isn’t allowed for function `A -> 1 | _ -> 0. In other words, the type needs to record the fact that `A takes no parameters.