[Haskell-beginners] Identical function and variable names and type inference

[Magnus Therning]

On Thu, Sep 3, 2009 at 8:35 AM, Magnus Therning<magnus at therning.org> wrote:
> On Wed, Sep 2, 2009 at 11:33 PM, Thomas Davie<tom.davie at gmail.com> wrote:
>> Ints can't make up the function part of an application, that must have type
>> (a -> b).
>
> Well, I don't think that's strictly true. foo::Int can be argued to be
> a function that takes no arguments and return an Int.  It is however
> very convenient to call such a function a "constant".  However, that
> such a function actully is constant is far from guaranteed in most
> programming languages.  It happens to be true in Haskell though, due
> to the strict separation between (proper) functions and "procedures"
> (functions with side effects, i.e. stuff in the IO monad).
>
>> In the mean time, the reason it didn't accept test id = id id is because it
>> must fix the argument id to only one type. It infers that id must have type
>> (a -> b), from the fact that it appears in the function position, then sees
>> it in the argument position, and infers that a = (a -> b) which obviously
>> causes a problem.  To resolve that problem, you need rank-2 polymorphism.
>
> Is that really correct?  I suspect the only thing that causes the OP
> problems is scoping.  If I understand you correctly I wouldn't be able
> to do the following:
>
> Prelude> let f x = x
> Prelude> :t f
> f :: t -> t
> Prelude> :t id
> id :: a -> a
> Prelude> :t id f
> id f :: t -> t
> Prelude> let g = id f
> Prelude> :t g
> g :: t -> t
> Prelude> id 5
> 5
> Prelude> f 5
> 5
> Prelude> g 5
> 5
>
> Clearly that's not a problem at all.

Ah, sorry, I just remembered that the 'forall' is implicit
(http://en.wikibooks.org/wiki/Haskell/Existentially_quantified_types).

/M

-- 
Magnus Therning                        (OpenPGP: 0xAB4DFBA4)
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