On 5/14/07, Roberto Zunino <[EMAIL PROTECTED]> wrote:
Also, using only rank-1:
polyf :: Int -> a -> Int
polyf x y = if x==0 then 0
else if x==1 then polyf (x-1) (\z->z)
else polyf (x-2) 3
Here passing both 3 and (\z->z) as y confuses the type inference.
Actually, I tried this in ghci... Should this work?
polyf.hs:
polyf x y = if x==0 then 0
else if x==1 then polyf (x-1) (\z->z)
else polyf (x-2) 3
NOTE: no type signature
Prelude> :l polyf
[1 of 1] Compiling Main ( polyf.hs, interpreted )
Ok, modules loaded: Main.
*Main> :t polyf
polyf :: forall a t1 t.
(Num (t1 -> t1), Num a, Num t) =>
a -> (t1 -> t1) -> t
The inference assigns y the type (t1 -> t1) even though it is assigned
the value 3?
Regards,
Chris
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
