无法匹配预期的类型Haskell

Question

我需要制作一个像[B,N,N]这样的例子列表，其中列表包含N的索引，如[1,2]，以及表示最终列表长度的int。

这是我的计划：

lignebn :: Int -> [Int] -> [Case] 
lignebn i [] = [] 
lignebn i x 
    | i == last x = [N] : lignebn (i-1) init x -- line 101 
    | otherwise = [B] : lignebn (i-1) init x 

下面是错误：

devoir.hs:101:17: 
    Couldn't match expected type ‘[Int]’ with actual type ‘Int’ 
    In the first argument of ‘last’, namely ‘(x)’ 
    In the second argument of ‘(==)’, namely ‘last (x)’ 

devoir.hs:101:29: 
    Couldn't match expected type ‘Int -> [Case]’ 
       with actual type ‘[Case]’ 
    The function ‘lignebn’ is applied to three arguments, 
    but its type ‘Int -> [Int] -> [Case]’ has only two 
    In the second argument of ‘(++)’, namely ‘lignebn (i - 1) init (x)’ 
    In the expression: [N] ++ lignebn (i - 1) init (x) 

devoir.hs:101:43: 
    Couldn't match expected type ‘[Int]’ 
       with actual type ‘[a0] -> [a0]’ 
    Probable cause: ‘init’ is applied to too few arguments 
    In the second argument of ‘lignebn’, namely ‘init’ 
    In the second argument of ‘(++)’, namely ‘lignebn (i - 1) init (x)’ 

devoir.hs:102:26: 
    Couldn't match expected type ‘Int -> [Case]’ 
       with actual type ‘[Case]’ 
    The function ‘lignebn’ is applied to three arguments, 
    but its type ‘Int -> [Int] -> [Case]’ has only two 
    In the second argument of ‘(++)’, namely ‘lignebn (i - 1) init (x)’ 
    In the expression: [B] ++ lignebn (i - 1) init (x) 

devoir.hs:102:40: 
    Couldn't match expected type ‘[Int]’ 
       with actual type ‘[a1] -> [a1]’ 
    Probable cause: ‘init’ is applied to too few arguments 
    In the second argument of ‘lignebn’, namely ‘init’ 
    In the second argument of ‘(++)’, namely ‘lignebn (i - 1) init (x)’ 
Failed, modules loaded: none. 

我不明白这是怎么回事，该代码似乎确定我...

Answer 1

在递归调用中，需要围绕init x的括号。事实上，您正在通过init函数和x。您可能还需要N和B列表，而不是列表中的列表，但很难肯定地说，因为您没有包含您的Case类型的定义。

lignebn :: Int -> [Int] -> [Case] 
lignebn i [] = [] 
lignebn i x 
    | i == last x = N : lignebn (i-1) (init x) -- line 101 
    | otherwise = B : lignebn (i-1) (init x) 

还要注意像last那部分功能被广泛气馁。更好的是这样的（未经测试）：

unsnoc :: [a] -> Maybe ([a], a) 
unsnoc [] = Nothing 
unsnoc (x : xs) = Just ps where 
    ps = case unsnoc xs of 
      Nothing -> ([], x) 
      Just (~(ys, y)) -> Just (x : ys, y) 

Answer 2

last期望列表，但是您已经在模式匹配中将其定义为并且Int

lignebn :: Int -> [Int] -> [Case] 
lignebn i (x:xs) = .... 

意味着

(x:xs)::[Int] 

这意味着x是一个Int。

---

在一个不相关的音符，在Haskell你不需要把函数调用前后的括号。不要像这样打电话给last(x)，而应该使用last x。它会工作，但不是惯用的哈斯克尔。