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

Oleg explained why those work in his last post. It's the exact same logic for each one.

f :: a -> a f x = x :: a

We explained that too: it's converted (alpha-converted, but I don't exactly know what 'alpha' refers to. I guess it's phase the type inferer goes through) to: f :: forall a. a -> a f x = x :: forall a1. a1 On one side, x has type a, on the other, it has type a1. Those are different polymorphic types, yet it's the same variable x hence the incompatibility. So it doesn't type-check. 2012/1/4 Thiago Negri

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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