為什么這樣的遞歸Haskell函數不起作用？

Question

這是代碼：

rep' :: Int -> a -> [a]
rep' 0 x = []
rep' n x = x:rep'(n-1, x)

我試圖這樣重寫它：

rep' :: Int -> a -> [a]
rep' 0 x = []
rep' n x = x:(rep' n-1 x)

但它也不起作用。

baby.hs:3:15-20: Couldn't match expected type ‘[a]’ with actual type ‘a0 -> [a0]’ …
    Relevant bindings include
      x :: a (bound at /Users/hanfeisun/Workspace/haskell/baby.hs:3:8)
      rep' :: Int -> a -> [a]
        (bound at /Users/hanfeisun/Workspace/haskell/baby.hs:2:1)
    Probable cause: ‘rep'’ is applied to too few arguments
    In the first argument of ‘(-)’, namely ‘rep' n’
    In the second argument of ‘(:)’, namely ‘(rep' n - 1 x)’
Compilation failed.
λ> 

有人對此有想法嗎？

Answer 1

Haskell在錯誤消息中表達其問題和期望，像這樣

Prelude> :{
Prelude|     let
Prelude|     {
Prelude|         rep' :: Int -> a -> [a];
Prelude|         rep' 0 x = [];
Prelude|         rep' n x = x:rep' (n-1, x);
Prelude|     }
Prelude| :}

<interactive>:73:22:
    Couldn't match expected type `[a]' with actual type `a0 -> [a0]'
    In the return type of a call of rep'
    Probable cause: rep' is applied to too few arguments
    In the second argument of `(:)', namely `rep' (n - 1, x)'
    In the expression: x : rep' (n - 1, x)

<interactive>:73:27:
    Couldn't match expected type `Int' with actual type `(Int, a)'
    In the first argument of rep', namely `(n - 1, x)'
    In the second argument of `(:)', namely `rep' (n - 1, x)'
    In the expression: x : rep' (n - 1, x)

在第一部分中

    Couldn't match expected type `[a]' with actual type `a0 -> [a0]'
    In the return type of a call of rep'
    Probable cause: rep' is applied to too few arguments

說，您已經將rep'的返回類型聲明為[a] ，但是它正在返回a0 -> [a0] ，這意味着它正在返回部分應用的函數。 可能的問題也提示給您

    Probable cause: rep' is applied to too few arguments

因此您可能向函數rep'傳遞了更少的參數。 在下一節中，

    Couldn't match expected type `Int' with actual type `(Int, a)'

表示希望有一個Int ，但得到了(Int, a) 。 在Haskell中，當您說(n-1, x) ，它將被視為元組對象，其中包含兩個元素。 因此，您實際上是在使用單個元組對象而不是兩個參數來調用rep' 。

要實際調用帶有兩個參數的rep' ，您可以這樣操作

rep' n x = x:rep' (n-1) x

現在，您正在使用兩個參數(n-1)和x調用rep' 。

Prelude> :{
Prelude|     let
Prelude|     {
Prelude|         rep' :: Int -> a -> [a];
Prelude|         rep' 0 x = [];
Prelude|         rep' n x = x:rep' (n-1) x;
Prelude|     }
Prelude| :}
Prelude> rep' 5 100
[100,100,100,100,100]

Answer 2

rep'的第一個參數應該是Int ，但是當您將其稱為rep' (n-1, x) ，第一個也是唯一的參數是元組。