On Mon, Oct 27, 2014 at 3:47 PM, Ovidiu Deac wrote:

Prelude> let f x = x * 2 Prelude> :t f f :: Num a => a -> a

The typeclass Num is defined like this:

class Num a where (+), (-), (*) :: a -> a -> a ...

...which means that the operator (*) expects two parameters of type a and returns a value of type a.

Since the definition of expr looks like this: Prelude> let f x = x * 2

...and 2 is an Int, I would expect that the type inferred for (*) is (Int -> Int -> Int) and thus f should be (Int -> Int)

Can somebody explain this?

Thanks!

Not sure of the terminology, but 2 can be coerced to any different type of Num, so that it can be used in non-Int expressions: 2 / 3 0.6666666666666666 Prelude> :t 2 / 3 2 / 3 :: Fractional a => a Prelude> :t 2 2 :: Num a => a Note that Fractional is also a type class, not a type like Int, so Floating types do the same thing. For that mater, if you enabled OverloadedStrings, you can get that behavior out of strings: Prelude> :t "abc" "abc" :: [Char] Prelude> :set -XOverloadedStrings Prelude> :t "abc" "abc" :: Data.String.IsString a => a