Use of forall as a sigil

[Vladislav Zavialov]

There is no *implicit* universal quantification in that example, but there
is an explicit quantifier. It is written as follows:

  forall a ->

which is entirely analogous to:

  forall a.

in all ways other than the additional requirement to instantiate the type
vatiable visibly at use sites.

- Vlad


On Thu, Dec 3, 2020, 19:12 Bryan Richter <b at chreekat.net> wrote:

> 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
>>
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