I haven't thought through the consequences of this fully, but it does seem plausible.
It says we must generalise over some ‘a’, but which ‘a’? What type does (\@a (\x. x)) have? I suppose you could say (forall b. (forall a. b -> b)), but presumably it could equally well have the type (forall a. a->a).
I think I can see an algorithm. But someone would need to write down a type system.
(I think I was probably was wrong to mutter about impredicativity.)
Simon
From: ghc-steering-committee On Behalf Of Richard Eisenberg
Sent: 05 March 2019 03:06
To: Eric Seidel
Cc: ghc-steering-committee@haskell.org
Subject: Re: [ghc-steering-committee] Discussion on #155 Type Variable in Labmdas
On Mar 3, 2019, at 8:44 PM, Eric Seidel mailto:eric@seidel.io> wrote:
Why not say that `\ @a -> body` infers the type `forall a. ty` where `body :: ty`, and let unification handle the rest?
That's intriguingly simple. I haven't thought through the consequences of this fully, but it does seem plausible. I'm not ready to commit to that, but after a few minutes' focused thought, I can't see any problems with it. In any case, the proposal as written is forward-compatible with such a change, as doing what you say will allow only more programs: ones accepted by the proposal will remain accepted and with the same meanings.
Richard
