It is a type level operator. You can parameterize things by it:
type Foo f = f Int String
foo :: Foo (->)
foo = show
-- Write instances for it:
class Profunctor p where
dimap :: (a -> b) -> (c -> d) -> p b c -> p a d
instance Profunctor (->) where
dimap ab cd bc = cd . bc . ab
-- Define aliases for it:
type Func a b = a -> b
map :: Func a b -> [a] -> [b]
-- or even,
map :: Func a b -> Func [a] [b]
-- Use it prefix
dropWhile :: ((->) a Bool) -> [a] -> [a]
-- Partially apply it:
instance MonadReader r ((->) r) where
ask = id
Matt Parsons
On Wed, Aug 15, 2018 at 10:53 PM, Paul
Hmm, very interesting question, really: what is the "->" in functions signatures: type-level operator or syntax only?
16.08.2018 01:20, Massimo Zaniboni wrote:
Il 15/08/2018 23:06, Stefan Chacko ha scritto:
3. Why do we use clinches in such definitions. I concluded you need
clinches if a function is not associative such as (a-b)+c . (Int->Int)->Int->Int
But also if a higher order function needs more than one argument. (a->b)->c .
Can you please explain it ?
funXYZ :: Int -> Int -> Int -> Int funXYZ x y z = (x - y) + z
if you rewrite in pure lamdda-calculus, without any syntax-sugars it became
fun_XYZ :: (Int -> (Int -> (Int -> Int))) fun_XYZ = \x -> \y -> \z -> (x - y) + z
so fun_XYZ is a function `\x -> ...` that accepts x, and return a function, that accepts a parameter y, and return a function, etc...
You can also rewrite as:
funX_YZ :: Int -> (Int -> (Int -> Int)) funX_YZ x = \y -> \z -> (x - y) + z
or
funXY_Z :: Int -> Int -> (Int -> Int) funXY_Z x y = \z -> (x - y) + z
and finally again in the original
funXYZ_ :: Int -> Int -> Int -> Int funXYZ_ x y z = (x - y) + z
I used different names only for clarity, but they are the same exact function in Haskell.
In lambda-calculus the form
\x y z -> (x - y) + z
is syntax sugar for
\x -> \y -> \z -> (x - y) + z
On the contrary (as Francesco said)
(Int -> Int) -> Int -> Int
is a completely different type respect
Int -> Int -> Int -> Int
In particular a function like
gHX :: (Int -> Int) -> Int -> Int gHX h x = h x
has 2 parameters, and not 3. The first parameter has type (Int -> Int), the second type Int, and then it returns an Int. Equivalent forms are:
g_HX :: (Int -> (Int -> Int)) g_HX = \h -> \x -> h x
gH_X :: (Int -> Int) -> (Int -> Int) gH_X h = \x -> h x
gHX :: (Int -> Int) -> Int -> Int gHX_ h x = h x
IMHO it is similar to logic: intuitively it seems easy and natural, but if you reflect too much, it is not easy anymore... but after you internalize some rules, it is easy and natural again.
Regards, Massimo _______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.
_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.