[Haskell-cafe] type inference question

[minh thu]

Thanks all!
Thu

2009/10/8 Lennart Augustsson <lennart at augustsson.net>:
> The reason a gets a single type is the monomorphism restriction (read
> the report).
> Using NoMonomorphismRestriction your example with a works fine.
>
> On Thu, Oct 8, 2009 at 12:29 PM, Cristiano Paris <frodo at theshire.org> wrote:
>> On Thu, Oct 8, 2009 at 11:04 AM, minh thu <noteed at gmail.com> wrote:
>>> Hi,
>>>
>>> I'd like to know what are the typing rules used in Haskell (98 is ok).
>>>
>>> Specifically, I'd like to know what makes
>>>
>>> let i = \x -> x in (i True, i 1)
>>>
>>> legal, and not
>>>
>>> let a = 1 in (a + (1 :: Int), a + (1.0 :: Float))
>>>
>>> Is it correct that polymorphic functions can be used polymorphically
>>> (in multiple places) while non-functions receive a monomorphic type ?
>>
>> First, "1" IS a constant function so it's in no way special and is a
>> value like any other.
>>
>> That said, the type of 1 is (Num t) => t, hence polymorphic. But, when
>> used in the first element of the tuple, a is assigned a more concrete
>> type (Int) which mismatches with the second element of the tuple,
>> which is a Float.
>>
>> If you swap the tuple, you'll find that the error reported by ghci is
>> the very same as before, except that the two types are swapped.
>>
>> Is it possible to rewrite the expression so as to work? The answer is
>> yes, using existential quantification (and Rank2 polymorphism).
>>
>> Here you are:
>> {-# LANGUAGE ExistentialQuantification, Rank2Types #-}
>> foo :: (forall a. (Num a) => a) -> (Int,Float)
>> foo = \x -> (x + (1 :: Int), x + (1 :: Float))
>>
>> Hence:
>>
>> foo 1 --> (2,2.0)
>>
>> Bye,
>>
>> Cristiano
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>