Referential transparency with polymorphism in Hask

问题:

Say I have a function:

f :: Int -> (Rational, Integer)
f b = ((toRational b)+1,(toInteger b)+1)

I want to abstract away the (+1) like so:

f :: Int -> (Rational, Integer)
f b = (h (toRational b)
      ,h (toInteger b))
    where h = (+1)

This wont work obviously, but if I specify the type signature it will work:

f :: Int -> (Rational, Integer)
f b = (h (toRational b)
      ,h (toInteger b))
    where h :: Num a => a -> a
          h = (+1)

Say I now want to further abstract the function by passing h as a parameter:

f :: Num a => Int -> (a -> a) -> (Rational, Integer)
f b g = (h (toRational b)
        ,h (toInteger b))
    where h :: Num a => a -> a
          h = g

I get an error that the inner a is not the same a as the outer one.

Does anyone know how to write this function correctly? I want to pass a polymorphic function g to f and use it polymorphically.

I have encountered this situation multiple times now in very different projects, and I could not find a good solution.

回答1:

I found the solution: using the forall quantifier like so:

{-# LANGUAGE RankNTypes #-}
f :: Int -> (forall a. Num a=> a -> a) -> (Rational, Integer)
f b g = (h (toRational b)
        ,h (toInteger b))
    where h :: Num a => a -> a
          h = g

Which of course can be turned into:

f :: Int -> (forall a. Num a=>a -> a) -> (Rational, Integer)
f b g = (g (toRational b)
        ,g (toInteger b))