Rank 2 polymorphism in pattern matching?
This counterintuitive typechecking result came up when I wrote a wrapper around runST. Is there some limitation of HM with respect to type checking pattern matching? data X a b = X (a -> a) run :: forall a. (forall b. X a b) -> a -> a -- This definition doesn't pass the typechecker run (X f) = f -- But this definition works run x = (\(X f) -> f) x
On 2006-04-08, C Rodrigues
This counterintuitive typechecking result came up when I wrote a wrapper around runST. Is there some limitation of HM with respect to type checking pattern matching?
data X a b = X (a -> a) run :: forall a. (forall b. X a b) -> a -> a -- This definition doesn't pass the typechecker run (X f) = f -- But this definition works run x = (\(X f) -> f) x
Have you tried run (X f) x = f x ? -- Aaron Denney -><-
Hello, See this message: http://article.gmane.org/gmane.comp.lang.haskell.general/13145/ Your (initial) program should work in GHC 6.2. I actually find this feature useful, but Simon apparently changed this when moving to GHC 6.4 and nobody complained... Apparently not many people use this feature. Cheers, Bruno On Sat, 08 Apr 2006 18:31:03 +0000, C Rodrigues wrote:
This counterintuitive typechecking result came up when I wrote a wrapper around runST. Is there some limitation of HM with respect to type checking pattern matching?
data X a b = X (a -> a) run :: forall a. (forall b. X a b) -> a -> a -- This definition doesn't pass the typechecker run (X f) = f -- But this definition works run x = (\(X f) -> f) x
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participants (3)
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Aaron Denney -
Bruno Oliveira -
C Rodrigues