Use of forall as a sigil

[Bryan Richter]

I must be confused, because it sounds like you are contradicting yourself.
:) In one sentence you say that there is no assumed universal
quantification going on, and in the next you say that the function does
indeed work for all types. Isn't that the definition of universal
quantification?

(We're definitely getting somewhere interesting!)

Den tors 3 dec. 2020 17:56Richard Eisenberg <rae at richarde.dev> skrev:

>
>
> On Dec 3, 2020, at 10:23 AM, Bryan Richter <b at chreekat.net> wrote:
>
> Consider `forall a -> a -> a`. There's still an implicit universal
> quantification that is assumed, right?
>
>
> No, there isn't, and I think this is the central point of confusion. A
> function of type `forall a -> a -> a` does work for all types `a`. So I
> think the keyword is appropriate. The only difference is that we must state
> what `a` is explicitly. I thus respectfully disagree with
>
> But somewhere, an author decided to reuse the same keyword to herald a
> type argument. It seems they stopped thinking about the meaning of the word
> itself, saw that it was syntactically in the right spot, and borrowed it to
> mean something else.
>
>
> Does this help clarify? And if it does, is there a place you can direct us
> to where the point could be made more clearly? I think you're far from the
> only one who has tripped here.
>
> Richard
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.haskell.org/pipermail/ghc-devs/attachments/20201203/35cbe076/attachment.html>