[Haskell-cafe] Deconstruction
haskell at kudling.de
haskell at kudling.de
Sat Dec 26 03:58:44 EST 2009
Hi,
while this works:
data Foo a = Foo a
unwrapFoo :: Foo a -> a
unwrapFoo (Foo x) = x
this:
{-# LANGUAGE ExistentialQuantification #-}
class BarLike a where
doSomething :: a -> Double
data Bar = forall a. BarLike a => Bar a
unwrapBar :: Bar -> a
unwrapBar (Bar x) = x
gives me:
Couldn't match expected type `a' against inferred type `a1'
`a' is a rigid type variable bound by
the type signature for `unwrapBar' at test.hs:8:20
`a1' is a rigid type variable bound by
the constructor `Bar' at test.hs:9:11
In the expression: x
In the definition of `unwrapBar': unwrapBar (Bar x) = x
How can i deconstruct the enclosed value of type a?
Thanks,
Lenny
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