Re: in defense of types

[Matthias Felleisen <address@hidden>]

Pixel, the point of Y is NOT to use let rec :)

For people who need to see a typed version in ML that works
for ALL types, see the last page of "The Little MLer". 

For those who need to see this in Java, take a look at the 
last page of "A Little Java, A Few Patterns."

-- Matthias


At 30 May 2002 12:56:10 +0200, Pixel wrote:
> Michael Vanier <mvanier@cs.caltech.edu> writes:
> 
> > > let z f = (fun x -> f (fun z -> x x z)) (fun x -> f (fun z -> x x z));;
> > > 
> > >   is not typable.  since two 'x' in 'x x z' cannot have the same type.
> > >   The key point is distinguish the type of these instances of 'x'.
> > > 
> > > # type 'a t = I of ('a t -> 'a);;
> > > type 'a t = I of ('a t -> 'a)
> > > # let z f = (function I x -> f (fun z -> x (I x) z))
> > >                 (I (function I x -> f (fun z -> x (I x) z)));;
> > > val z : (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b = <fun>
> > > 
> > >   voila! you can type it. Now, let's try simple example.
> >
> > This is a brilliant and non-obvious demonstration; thanks a lot!  You saved
> > me hours of scratching my head ;-)
> > 
> > BTW I'm not sure what you mean when you say that Y is not usable in eager
> > languages.  You can define Y in scheme, which is eager.  Do you mean in
> > eager statically-typed languages?
> 
> I don't understand either. And isn't this much simpler and working:
> 
> let rec y f e = f (y f) e
> let ffact f n = if n = 1 then 1 else n * f (n - 1)
> ;;
> y ffact 5
> 
> ??