[Haskell-beginners] map type explanation

[Lawrence Bottorff]

So in effect

a -> b -> [a] -> [b]

wants to be, would be

a -> (b -> ([a] -> [b]))

without the parens (which is a natural result of lambda calculus, perhaps?)
-- which is not what is meant by map. But underlying a Haskell type
declaration is currying, is it not? At the type declaration level, it's all
currying, correct?

Conceptually, I understand how the a -> b "event" needs to be a "package"
to apply to the list [a]. The map function commandeers the target function
(which alone by itself does some a -> b evaluation) to be a new object that
is then applied to each member of list [a]. Good. So (a -> b) then is a
notation that signifies this "package-ness".

Does anyone have examples of other "packaging" where a function doing some a
-> b is changed to (a -> b) ?

On Fri, Dec 18, 2020 at 5:18 PM Bruno Barbier <brubar.cs at gmail.com> wrote:

>
> Hi Lawrence,
>
> Lawrence Bottorff <borgauf at gmail.com> writes:
>
> > Why is it not just
> >
> > a -> b -> [a] -> [b]
> >
> > again, why the parentheses?
>
> In Haskell, (->) is a binary operator and is right associative. If you
> write:
>
>    a -> b -> [a] -> [b]
>
> it implicitly means:
>
>    a -> (b -> ([a] -> [b]))
>
> So here, you need explicit parenthesis:
>
>    (a -> b) -> [a] -> [b]
>
> to mean:
>    (a -> b) -> ([a] -> [b])
>
> It's more about parsing binary operators than about types.
>
> Does it help ?
>
> Bruno
>
> > On Fri, Dec 18, 2020 at 4:10 PM Ut Primum <utprimum at gmail.com> wrote:
> >
> >> Hi,
> >>
> >> a -> b  is the type of a function taking arguments of a generic type (we
> >> call it a) and returning results of another type, that we call b.
> >>
> >> So
> >> (a -> b ) -> [a] -> [b]
> >> Means that you have a first argument that is a function (a-> b),  a
> second
> >> argument that is a list of elements of the same type of the function
> input,
> >> and that the returned element is a list of things of the type of the
> output
> >> of the function.
> >>
> >> Cheers,
> >> Ut
> >>
> >> Il ven 18 dic 2020, 23:02 Lawrence Bottorff <borgauf at gmail.com> ha
> >> scritto:
> >>
> >>> Thank you, but why in
> >>>
> >>> map :: (a -> b) -> [a] -> [b]
> >>>
> >>> are there parentheses around a -> b ? In general, what is the currying
> >>> aspect of this?
> >>>
> >>>
> >>> On Fri, Dec 18, 2020 at 12:43 PM David McBride <toad3k at gmail.com>
> wrote:
> >>>
> >>>> They are not parameters, they are the types of the parameters.
> >>>>
> >>>> In this case a can really be anything, Int, Char, whatever, so long as
> >>>> the function takes a single argument of that type and the list that is
> >>>> given has elements of that same type.
> >>>> It is the same for b, it doesn't matter what b ends up being, so long
> as
> >>>> when you call that function the function's return value is compatible
> with
> >>>> the element type of the list that you intended to return from the
> entire
> >>>> statement.
> >>>>
> >>>> You can mess with it yourself in ghci to see how type inference works.
> >>>>
> >>>> >:t show
> >>>> :show :: Show a => a -> String
> >>>> >:t map show
> >>>> map show :: Show a => [a] -> [String]
> >>>> > :t flip map [1::Int]
> >>>> > flip map [1::Int] :: (Int -> b) -> [b]
> >>>>
> >>>>
> >>>> On Fri, Dec 18, 2020 at 1:31 PM Lawrence Bottorff <borgauf at gmail.com>
> >>>> wrote:
> >>>>
> >>>>> I'm looking at this
> >>>>>
> >>>>> ghci> :type map
> >>>>> map :: (a -> b) -> [a] -> [b]
> >>>>>
> >>>>> and wondering what the (a -> b) part is about. map takes a function
> >>>>> and applies it to an incoming list. Good. Understood. I'm guessing
> that the
> >>>>> whole Haskell type declaration idea is based on currying, and I do
> >>>>> understand how the (a -> b) part "takes" an incoming list, [a] and
> >>>>> produces the [b] output. Also, I don't understand a and b very well
> >>>>> either. Typically, a is just a generic variable, then b is another
> >>>>> generic variable not necessarily the same as a. But how are they
> being
> >>>>> used in this type declaration?
> >>>>>
> >>>>> LB
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