Hi @pelotom, thanks for sharing.

How do you define a Monad instance for Either?

How do you define a Monad instance for Either?

Good question. The only thing I can think to do is have a parameterized instance creator

const RightMonad = <L>() => Monad<Either<Fixed<L>, _>>({
  pure: right,
  bind: (ma, f) => (ma.tag === "left" ? ma : f(ma.right)),
});

const either: Either<string, number> = right(42);
const result = RightMonad<string>().map(either, n => n + 1);
expect(result).toEqual(right(43));

The new Fixed operator is needed to instruct $ to leave bound type parameters unmolested. What do you think, is there a better approach?

It"s a shame that you can"t just say:

type EitherMonad = <L> Monad<Either<Fixed<L>, _>>

The ability to introduce foralls is restricted to functions.

Actually it looks like the Fixed operator is no longer needed here, and we can just write

const RightMonad = <L>() => Monad<Either<L, _>>({
  pure: right,
  bind: (ma, f) => (ma.tag === "left" ? ma : f(ma.right)),
});

Not really sure why this is 😅

@masaeedu

It"s a shame that you can"t just say:

type EitherMonad = <L> Monad<Either<Fixed<L>, _>>

The ability to introduce foralls is restricted to functions.

Yes, it is... I hope your issue microsoft/TypeScript#17574 gets some attention!