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 <[email protected]>:
>> 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 <[email protected]>
>
> On Wed, Jan 4, 2012 at 08:41, Yves Parès <[email protected]> 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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