[Haskell-beginners] Fwd: Re: Multiple letters between -> ->

[Marcus Manning]

Sorry again fo the long break,

and thanks for explanation.

Three Questions:

1.) Because (->) is defined with itself, it is also a Meta Type 
Constructor (TYPE == variable Rank Type == a Meta Type ?)

2.) How do I define a function which takes a 3-Kinded Type:

let f:: h (g a); f a = a

where g * -> * and h (*->*)->*, but it did not work as:

h is deduced to be h * -> *.

Prelude> let f:: h (g a); f a = a

<interactive>:42:18: error:
     • Couldn't match type ‘h’ with ‘(->) (g a)’
       ‘h’ is a rigid type variable bound by
         the type signature for:
           f :: forall (h :: * -> *) (g :: * -> *) a. h (g a)
         at <interactive>:42:5-15
       Expected type: h (g a)
         Actual type: g a -> g a
     • The equation(s) for ‘f’ have one argument,
       but its type ‘h (g a)’ has none
     • Relevant bindings include
         f :: h (g a) (bound at <interactive>:42:18)

What can I do?

3.)
Is there any tool in Haskel where I can see which function parameter is 
bound to the given input, for instance something like a function showBinds:

showBinds const (\c -> "s") (\(b,c) -> "c")
Prelude> Param a gets (\c -> "s")
                            b gets (\(b,c) -> "c")

This is a simple example, but it gets more complicated with massive use 
of autocurrying.




On 11/25/2017 05:39 PM, Francesco Ariis wrote:
> On Sat, Nov 25, 2017 at 04:19:04PM +0100, Marcus Manning wrote:
>> But why I can call g with Just:
>>
>>
>> let g :: h a b -> h a b; g a = a
>>
>> g Just
>>
>> but Just is a->Maybe a
> Because (->) is a type constructor itself, just with
> convenient infix syntax:
>
>      λ> :k (->)
>      (->) :: TYPE q -> TYPE r -> *
>
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