Use of forall as a sigil

[Andrey Mokhov]

Hi John,

> - We are already getting `forall {a}.`, so it fits nicely with that.

Interesting, I wasn't aware of this. Could you point me to the relevant proposal?

> - However, it would have to be `forall @a ->`, 

Oh, that seems even worse than `forall a ->` to me.

> because `forall a.` is already an invisible quantification,
> unless one wants to just change the meaning of `forall a.`!

I'm confused. I wasn't suggesting to change the meaning of `forall a.`.

My suggestion was pretty incremental:

* `forall a.` stays as is: it allows for both invisible and visible type arguments.
* `forall @a.` requires a visible type argument.

Cheers,
Andrey

-----Original Message-----
From: John Ericson [mailto:john.ericson at obsidian.systems] 
Sent: 22 November 2020 16:41
To: Andrey Mokhov <andrey.mokhov at newcastle.ac.uk>; Richard Eisenberg <rae at richarde.dev>
Cc: ghc-devs at haskell.org
Subject: Re: Use of forall as a sigil


I have thought about this too, and don't believe it has been widely
discussed.

- We are already getting `forall {a}.`, so it fits nicely with that.

- However, it would have to be `forall @a ->`, because `forall a.` is
already an invisible quantification, unless one wants to just change the
meaning of `forall a.`!

John