Re: [Haskell-cafe] Solved but strange error in type inference

Do not compile: f :: a -> a f x = x :: a Couldn't match type `a' with `a1' `a' is a rigid type variable bound by the type signature for f :: a -> a at C:\teste.hs:4:1 `a1' is a rigid type variable bound by an expression type signature: a1 at C:\teste.hs:4:7 In the expression: x :: a In an equation for `f': f x = x :: a Any of these compiles: f :: a -> a f x = undefined :: a f :: Num a => a -> a f x = undefined :: a f :: Int -> Int f x = undefined :: a f :: Int -> Int f x = 3 :: (Num a => a) Can someone explain case by case? Thanks, Thiago. 2012/1/4 Yves Parès :

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I don't see the point in universally quantifying over types which are already present in the environment

I think it reduces the indeterminacy you come across when you read your program ("where does this type variable come from, btw?")

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So is there anyway to "force" the scoping of variables, so that f :: a -> a f x = x :: a becomes valid?

You mean either than compiling with ScopedTypeVariables and adding the explicit forall a. on f? I don't think.

2012/1/4 Brandon Allbery

On Wed, Jan 4, 2012 at 08:41, Yves Parès wrote:

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Would you try:

f :: a -> a

f x = undefined :: a

And tell me if it works? IMO it doesn't.

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  It won't

Apparently, Yucheng says it does.

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