Ok, a and s are type variables. It may be common
usage for computer people but for someone with a math background it is difficult
to call forall a a type . In logic one expects a statement in
the scope of forall a like this
(forall a) statement. The statement may or may not depend on
a.
So let's come backe to runST.
runST ::( forall s) (ST s a )->
a
So I insist what is the statement
ST s a ?
Scott
----- Original Message -----
Sent: Monday, August 19, 2002 4:26 AM
Subject: Re: Question aboutthe use of an inner
forall
> At 2002-08-18 18:19, Scott J. wrote:
>
> >A
question: s is not a type variable as a isn't it?
>
> Both are
quantified type variables. But only one of those quantifiers can
> be
placed at the top level of the expression.
>
> runST
:: forall a. ((forall s. ST s a) -> a)
>
> > I mean a can be
of type Integer while s cannot.
>
> Well yes, you can specialise
"forall a" to Integer. So runST can be
> specialised to this
type:
>
> runST ::[special] (forall s. ST s Integer)
-> Integer
>
> ...whereas you cannot specialise "s" in the same
way. Think of it in
> terms of logic, if for all a: f(a), then by simple
specialisation
> f(Integer).
>
> --
> Ashley Yakeley,
Seattle WA
>
>