努力理解“无法构造无限类型”错误

Question

我正在尝试解决 HackerRank 问题，但遇到了一个我无法弄清楚的错误。问题出在solve1函数上。确切的错误是：cannot construct the infinite type a ~ t0 a. Expected type ([t0 a], [t0 a], [t0 a]) Actual type ([a], [a], [a]). In the second argument of `tripleMap`, namely '(tripList xs)'。

我一直看着这些类型，它们在我眼中继续显得正确。tripList获取数字列表并返回数字列表的三元组。tripleMap将一组数字列表作为其第二个参数。

在tripList我的 REPL 中进行测试时，我得到了想要的结果：

> tripList [1,0,-1,0,1] 
([1,1],[0,0],[-1])

这是我的代码：

length' :: (Foldable t, Num b, Fractional b, Ord b) => t a -> b
length' = foldr (\_ acc -> 1 + acc) 0

tripList :: (Num a, Ord a) => [a] -> ([a], [a], [a])
tripList xs =
  ( filter (>0) xs
  , filter (==0) xs
  , filter (<0) xs )

foldSolution :: (Foldable t, Num a, Ord a, Fractional a)
             => a -> t a -> a
foldSolution n = foldr (\x y -> x/n + y/n) 0

tripleMap :: (a -> b) -> ([a], [a], [a]) -> ([b], [b], [b])
tripleMap f (a, b, c) = (map f a, map f b, map f c)

solve1 :: (Num a, Ord a) => [a] -> ([a], [a], [a])
solve1 xs = tripleMap (foldSolution (length' xs)) (tripList xs)

Answer 1

你foldSolution (length' xs)需要t a和回报a。所以，当你打开tripleMap它tripList xs（它有 type ([a], [a], [a])）时，你会得到一个 type 的元组(a, a, a)， not ([a], [a], [a])，这就是你得到错误的原因。