[Haskell-cafe] (Un)termination of overloading resolution
sulzmann at comp.nus.edu.sg
Wed Feb 22 00:24:03 EST 2006
oleg at pobox.com writes:
> > Let's consider the general case (which I didn't describe in my earlier
> > email).
> > In case we have an n-ary type function T (or (n+1)-ary type class
> > constraint T) the conditions says for each
> > type T t1 ... tn = t
> > (or rule T t1 ... tn x ==> t)
> > then rank(ti) > rank(t) for each i=1,..,n
> I didn't know what condition you meant for the general form. But the
> condition above is not sufficient either, as a trivial modification of the
> example shows. The only modification is
> instance E ((->) (m ())) (() -> ()) (m ()) where
> test = foo (undefined::((() -> ()) -> ()) -> ()) (\f -> (f ()) :: ())
> Now we have t1 = ((->) (m ())) : two constructors, one variable
> t2 = () -> () : three constructors
> t = m () : one constructor, one variable
> and yet GHC 6.4.1 loops in the typechecking phase as before.
rank (() ->()) > rank (m ()) does NOT hold.
Sorry, I left out the precise definition of the rank function
in my previous email. Here's the formal definition.
rank(x) is some positive number for variable x
rank(F t1 ... tn) = 1 + rank t1 + ... + rank tn
where F is an n-ary type constructor.
rank (f t) = rank f + rank t
f is a functor variable
Hence, rank (()->()) = 3
rank (m ()) = rank m + 1
We cannot verify that 3 > rank m + 1.
So, I still claim my conjecture is correct.
Oleg, can you next time please provide more details
why type inference does not terminate. This will help
others to follow our discussion.
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