我如何解决这个问题：One:: Num a => a -> Bool -> a in Haskell?

Question

Define functions that have the following type signature!

One :: Num a => a -> Bool -> a

One = ?

I do not understand how a Num can be a bool and become again a Num. Could anybody help for me to understand and solve it.

Answer 1

how a Num can be a bool and become again a Num. Could anybody help for me to understand this.

Well, this would be the case if the grouping in the type was to the left, like so

one :: Num a => (a -> Bool) -> a

But actually, the grouping is to the right:

one :: Num a => a -> (Bool -> a)

So we need to find a way to turn an a type value into a way to turn a Bool value to an a type value.

Since we will have had this a type value, we could just return it:

one a = two 
  where
  two b = a

Do note that I had to change One to one . While One is syntactically a type in Haskell, one is a value which has a type. Which you show.

So what we did here is define the value one , which has this type. Or in Haskell,

one :: a -> (Bool -> a)

But wait, where did the Num a => constraint go? We didn't use it at all, and that's why all we could do was to just return the a .

But what does it mean, for a to be a Num ?

GHCi> :i Num

class Num a where
  (+) :: a -> a -> a
  (*) :: a -> a -> a
  (-) :: a -> a -> a
  negate :: a -> a
  abs :: a -> a
  signum :: a -> a

So then, any Num has absolute value, and can be negated. Thus a possible definition is

one :: Num a => a -> (Bool -> a)
one a = two
  where
  two b | b      = negate a
        | otherwise = abs a