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. That is 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.
+ 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.