[Haskell-beginners] $ versus .

[Lawrence Bottorff]

> :t (init tail)
: error:
:     * Couldn't match expected type `[a]'
:                   with actual type `[a0] -> [a0]'
:     * Probable cause: `tail' is applied to too few arguments
:       In the first argument of `init', namely `tail'
:       In the expression: (init tail)

> :t (init . tail)
: (init . tail) :: [a] -> [a]

> :t init $ tail
: error:
    * Couldn't match expected type `[a]'
                  with actual type `[a0] -> [a0]'
    * Probable cause: `tail' is applied to too few arguments
      In the second argument of `($)', namely `tail'
      In the expression: init $ tail

> chopEnds = init $ tail
> chopEnds [1,2,3]
error: ...
    * Variable not in scope: chopEnds1 :: [Integer] -> t
...

but then

> init $ tail [1,2,3]
[2]

Not sure what I'm missing here. It doesn't make sense to me that the last
expression works, but no version of a closure

chopEnds = init $ tail

does.

On Mon, Jan 25, 2021 at 7:58 PM Kim-Ee Yeoh <ky3 at atamo.com> wrote:

> init $ tail [1,2,3]
> = init (tail ([1,2,3])) -- a la Lisp
>
> Now, functional programming is awesomest at abstractions. What if we could
> abstract out "init (tail"?
>
> Then we could write
>
> chopEnds = init (tail
>
> But that looks weird. It's only got the left half of a parens pair!
>
> Does that explain why you should not expect the same result?
>
> A separate question is why the compiler even type-checks "init $ tail" in
> the first place. What do you think is going on there?
>
> On Tue, Jan 26, 2021 at 1:16 AM Lawrence Bottorff <borgauf at gmail.com>
> wrote:
>
>> I've got this
>>
>> > init $ tail [1,2,3]
>> [2]
>>
>> and this
>>
>> > chopEnds = init $ tail
>> > chopEnds [1,2,3]
>> [1,2]
>>
>> What happened? Why is it not just init $ tail [1,2,3] ?
>>
>> This works fine
>>
>> > chopEnds2 = init . tail
>> > chopEnds2 [1,2,3]
>> [2]
>>
>> What am I missing?
>>
>> LB
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> --
> -- Kim-Ee
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