Return specific type within Haskell

问题:

I have a pretty general question about Haskell's type system. I'm trying to become familiar with it, and I have the following function:

getN :: Num a => a
getN = 5.0 :: Double

When I try to compile this, I get the following error:

Couldn't match expected type `a' against inferred type `Double'
  `a' is a rigid type variable bound by
      the type signature for `getN' at Perlin.hs:15:12
In the expression: 5.0 :: Double
In the definition of `getN': getN = 5.0 :: Double

As I understand this, the function is set up to "return" a type in the class Num. Double is in this class (http://www.zvon.org/other/haskell/Outputprelude/Num_c.html), so I would have expected that it would be okay to "return" a Double in this case.

Can someone explain this please?

回答1:

A function with signature Num a => a is expected to work for any type in the class Num. The implementation 5.0 :: Double just works for one type, not for all types of the class, so the compiler complains.

An example of a generic function would be:

square :: (Num a) => a -> a
square x = x * x

This works for any type that is a Num. It works for doubles, integers and whatever other numbers you want to use. Because of that it can have a generic type signature that just requires the parameter to be in class Num. (Type class Num is necessary because the function uses multiplication with *, which is defined there)

回答2:

To add to sth's answer: Haskell is not object-oriented. It's not true that Double is a subclass of Num, so you cannot return a Double if you promise to return a polymorphic Num value, like you can in, say, Java.

When you write getN :: Num a => a you promise to return a value that is fully polymorphic within the Num constraint. Effectively this means that you can only use functions from the Num type class, such as +, *, - and fromInteger.

回答3:

Check out Existentially quantified types.

One way to solve it would be to define a new data type

data NumBox = forall n. Num n => NumBox n

You'll need -XExistentialQuantification to get this to work.

Now you can write something like

getN :: NumBox
getN = NumBox (5.0 :: Double)

You can also define a NumBox-list as

let n3 = [NumBox (4.0 :: Double), NumBox (1 :: Integer), NumBox (1 :: Int) ]