CODEWITHSUNDEEP
functionhaskelloperatorsfunction-calls
user634615
user634615

Reputation: 677

Use of function application operator in Haskell

What does the following expression means in haskell?

($ 3)

ghci shows the following type

($ 3) :: Num a => (a -> b) -> b.

Upvotes: 3

Views: 195

Answers (3)

Viktor Mellgren
Viktor Mellgren

Reputation: 4506

Another way to look at it is 

($) :: (a -> b) -> a -> b
3 :: Num a => a

and when you "insert 3" in the ($) it will become

($ 3) :: Num a => (a -> b) -> b.

due to that you no longer need to supply the a, but the function you need to supply is now restricted to num, since the 3 can be any numeric type.

This is at least how I look at functions in Haskell, like substitution in algebra.

Upvotes: 3

chi
chi

Reputation: 116139

($ 3) is a section, and is equivalent to \f -> f 3, which takes a function argument and applies it to 3.

If we considered 3 to be an integer, we would have that the type of f is Int -> b (for any b), so the type of ($ 3) would be (Int -> b) -> b.

Things in Haskell are a bit more complex, since 3 can be of any numeric type, so we don't really need f :: Int -> b, it's enough if f :: a -> b where a is a numeric type.

Hence we get ($ 3) :: Num a => (a -> b) -> b.

Upvotes: 8

user1804599
user1804599

Reputation:

(@ x) for any operator @ is equivalent to \a -> a @ x; so ($ 3) is equivalent to \f -> f $ 3, i.e. a function that applies any function you pass it to 3. This syntax is called "sections".

> let f = ($ 3)
> f show
"3"
> f square
9

Upvotes: 6

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