CODEWITHSUNDEEP
haskell
Tomer
Tomer

Reputation: 1229

Order of evaluation and type check with ($)

I'm trying to learn how the ($) operator works. I run 

(+5) ($)  7

I get 

 * Non type-variable argument in the constraint: Num (a -> b)
      (Use FlexibleContexts to permit this)
    * When checking the inferred type
        it :: forall a b.
              (Num (a -> b), Num ((a -> b) -> a -> b)) =>
              a -> b

Could anyone help me understand why I get this error ?

Upvotes: 1

Views: 118

Answers (2)

Mihalis
Mihalis

Reputation: 356

Much like (+) is the prefix form of +.

($), called function application, is the prefix form of $.

> (+) 1 2 == 1 + 2
True

So if you want to apply (+5) to 7 then ($) would do that in this syntax

> ($) (+5) 7
12

which is equivalent to

> (+5) $ 7
12

Note that $ is most often used to simplify syntax.

Upvotes: 2

Tomer
Tomer

Reputation: 1229

I should have done 

(+5) $  7

I still not sure I understand the difference.

Upvotes: 0

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