haskell functions declaration

Question

I am having problem understanding signatures/arguments/inputs of Haskell functions. Before anyone complains, yes I have done my research but can not seem to find a good answer or explanation.

basically I have a function :

update :: Eq a => (a->b) -> b -> a -> (a->b)

How do I make sense of (a->b) -> b -> a -> (a->b)? I see it as taking an input of a function followed by 2 values and outputting a function??

I have 2 different functions which do same thing, one which uses 3 arguments and one with 4 but the header (arguments of function) is same.

(1)

update :: Eq a => (a->b) -> b -> a -> (a->b)
update s v x y = if x==y then v else s y

(2)

update :: Eq a => (a->b) -> b -> a -> (a->b)
update s v y = ss
  where ss x = if ( x == y )
                 then v
                 else s x

They both compute same thing but I do not understand why (1) uses 4 inputs "update s v x y" and (2) "update s v y" uses 3 inputs.

Any help will be greatly appreciated.

Did you not come across currying in your research? (1) is a function with 4 arguments returning a value of type b. (2) is a functions with 3 arguments returning a function with 1 argument returning a value of type b. But actually Haskell only has functions with one argument. So both (1) and (2) are functions returning a function returning a function returning a function returning a value of type b (which might be a function, of course). — Chris
– Chris, Commented Apr 11, 2014 at 14:27
I'd say that your update (1) is (a -> b) -> b -> a -> a -> b, so it's visibly a function of 4 arguments; ghci's :t also says so. Removing the parens around the last a -> b is just currying. — 9000
– 9000, Commented Apr 11, 2014 at 14:30

daniel gratzer · Accepted Answer · 2014-04-11 14:24:07Z

6

Welcome to the wonderful world of currying.

Let's start with a simple function

allEq :: Int -> Int -> Int -> Bool
allEq x y z = x == y && y == z

Now in Haskell -> is right associative, so this is

allEq :: Int -> (Int -> (Int -> Bool)))
allEq x y z = x == y && y == z

allEq x y = \z -> x == y && y == z
allEq x = \y -> \z -> x == y && y == z
allEq = \x -> \y -> \z -> x == y && y == z

Now, in Haskell we know that we can "lift" an expression into a let or where binding without changing it.

 allEq x y = foo
   where foo z = x == y && y == z

And this is how we get to your function.

answered Apr 11, 2014 at 14:24

daniel gratzer

54k11 gold badges100 silver badges137 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Shayan C Over a year ago

Great answer, thank you. Just one thing, when I do this outside a function and in console, let allEq x = \y -> \z -> x == y && y == z, it works as it should but let alleq = \x -> \y -> \z -> x == y && y == z gives an error, why is this?

daniel gratzer Over a year ago

@user2371066 That's entirely unrelated to something called "the monomorphism restriction".

Shayan C Over a year ago

one last thing if you are free, as I asked above, I think (a->b) -> b -> a -> (a->b) is taking in a function and 2 values and outputting a function, then why do we have 4 values "update s v x y"? how do I read this correctly?

daniel gratzer Over a year ago

@user2371066 Remember that since it's right assocative it's just (a -> b) -> b -> a -> a -> b

Collectives™ on Stack Overflow

haskell functions declaration

1 Answer 1

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related