另一个无限类型错误

Question

I'm trying to create a function that composes a function f with a function g that takes multiple arguments.

c :: (a -> b) -> (c -> a) -> c -> b
c x y z = x(y(z))
lift = c c

(lift $ lift $ lift ... c) creates the desired function:

*Main> (lift $ lift $ lift $ lift $ lift $ c) (\x -> x+2) (\x y z a b c -> x*y*z-(a*b*c)) 1 2 3 4 5 6
-112

However, when I attempt to define a function to produce a function that takes f, g, then n arguments (to save typing), the following error occurs:

cn 1 = c
cn k = lift (cn(k-1))
<interactive>:9:1: error:
    * Occurs check: cannot construct the infinite type: a ~ c0 -> a
      Expected type: t -> (a -> b) -> a -> b
        Actual type: t -> (a -> b) -> (c0 -> a) -> c0 -> b
    * Relevant bindings include
        cn :: t -> (a -> b) -> a -> b (bound at <interactive>:9:1)

Why does this error occur, and how might I resolve it?

Answer 1

You cannot make polyvariadic functions in Haskell without typeclasses.

When you try to make a function that takes either one or two arguments, you try to make a function of type a -> b and a -> p -> b . Hence, the compiler must infer that the type b is equivalent to p -> b , so the type of the function becomes a -> p -> p -> p -> ... . In other words, an infinite type.

A similar problem occurs when you try to make this function. The first line is fine:

cn 1 = c

This would imply something like cn :: Int -> (b -> c) -> (a -> b) -> a -> c . However, we now have a problem in the second line:

cn k = lift (cn (k-1))

Since we know that cn :: Int -> (b -> c) -> (a -> b) -> a -> c , we must infer that cn (k-1) :: (b -> c) -> (a -> b) -> a -> c . However, since lift :: (x -> y -> z) -> x -> (w -> y) -> w -> z , we see that the return value must be of type (b -> c) -> (w -> a -> b) -> w -> c , which clashes with the original type declaration.

TL;DR: you can't change the type of a Haskell function/value via values. This means you can't write polyvariadic functions without certain tricks.