Re: Use of forall as a sigil

For me, the problem with `forall @a .` is that it seems to go in the wrong direction: parameters declared with @ must not have @ when passed, and parameters declared without @ must have a @ if the parameter is passed explicitly. However, if we say that @ makes a thing that is normally implicit explicit, maybe it works? Richard

On Nov 22, 2020, at 3:03 PM, Andrey Mokhov wrote:

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@obsidian.systems] Sent: 22 November 2020 16:41 To: Andrey Mokhov ; Richard Eisenberg Cc: ghc-devs@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