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.
Do not compile:
Couldn't match type `a' with `a1'
f :: a -> a
f x = x :: a
`a' is a rigid type variable bound bythe 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 :: aIn an equation for `f': f x = x :: a
Any of these compiles:
f :: Num a => a -> a
f :: a -> a
f x = undefined :: a
f x = undefined :: af :: Int -> Int
f x = undefined :: af :: Int -> Int
f x = 3 :: (Num a => a)
Can someone explain case by case?
Thanks,
Thiago.
2012/1/4 Yves Parès <limestrael@gmail.com>:
>> 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?")
>
>
>> 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 <allbery.b@gmail.com>
>
> On Wed, Jan 4, 2012 at 08:41, Yves Parès <limestrael@gmail.com> wrote:
>>
>> Would you try:
>>
>> f :: a -> a
>>
>> f x = undefined :: a
>>
>> And tell me if it works? IMO it doesn't.
>
>> It won't
>
> Apparently, Yucheng says it does.
>
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