Question

为什么以下haskell函数会出错？

目的是在列表中复制给定元素k次并返回列表。

功能：

--Replicate the element given number of times and return the list.
replicateElement :: (Integral k) => a -> k -> [a]
replicateElement x 0 = error "Cannot replicate Zero times"
replicateElement x 1 = [x]
replicateElement x k = [x] ++ replicateElement(x k-1)

错误：

Couldn't match expected type `[a]' with actual type `k0 -> [a0]'
    In the return type of a call of `replicateElement'
    Probable cause: `replicateElement' is applied to too few arguments
    In the second argument of `(++)', namely
      `replicateElement (x k - 1)'
    In the expression: [x] ++ replicateElement (x k - 1)
Failed, modules loaded: none.

Answer 1

写作时

replicateElement(x k-1)

您只将一个参数传递给replicateElement，即(x k-1)，这也意味着((x k) - 1)，而您实际上并不打算让x成为一个函数。相反，你应该写的是：

replicateElement x (k-1)

Answer 2

你的最后一行应该是：

replicateElement x k = [x] ++ replicateElement x (k-1)

您不需要在Haskell中为函数参数添加括号。

模式匹配功能给出错误

2 个答案: