Haskell列表反转错误 [英] Haskell List Reversal Error

问题描述

我已经知道列表反转，并且导致了以下代码：

p>

  myreverse list1 
 | list1 == [] = list1 
 |否则=（myreverse（尾列表1））:(头列表1）
  

不幸的是，上面的代码导致以下错误：

 发生检查：无法构造无限类型：a = [[a]] 
预期类型： [[a]] 
推断类型：a 
在'（:)'的第二个参数中，即'（head list1）'
在表达式中：（myreverse（tail list1）） ：（head list1）
  

PS：当我在代码片段上运行时，出现同样的错误我写下mylast编码如下：

  mylast list 
 | list == [] = [] 
 |否则= mylast_helper（头部列表1）（尾部列表1）
 
 mylast_helper项目列表2 
 | list2 == [] = item 
 |否则= mylast_helper（head list2）（tail list2）
  

在递归辅助函数的情况下发生错误。

编辑：感谢所有的输入，我想我忘了提及问题的约束禁止使用++运算符。我会记住我今后创建的问题。

干杯，
-Zigu

解决方案

您正在使用函数

 （:) :: a  - > [a]  - > [b] 
  

带有不正确的参数：

  myReverse（tail list1）:: [a] 
 head list1 :: a 
  

在你的函数中，列表 list1 必须有 a 类型。因此，第二个参数 必须具有类型 [a] 。 GHC警告你它不能构建你指定的类型。列表的头部在结构上比列表的尾部小，但是您告诉它列表的头部具有类型 [a] ，但列表的尾部具有类型 a 。

然而，如果你仔细观察你的类型，你会注意到你可以将的头添加使用（++）递归调用 myreverse ：> list1 ：

  myReverse xs = case（null xs）of 
 True  - > xs 
 False  - > 
  

这里，

  [head xs] :: [a] 
 myReverse（tail xs）:: [a] 
  

符合追加的类型：

（++）:: [a] - > [a] - >然而，实现 reverse 有很多更好的方法。 Prelude将 reverse 定义为左侧折叠（。 reverse 的另一个版本可以使用右侧折叠实现，与您的 myReverse 非常相似。函数：
$ b $ pre $ reverse xs = foldr（\ x xs - > xs ++ [x]）[] xs





I'm writing a list reversal program for haskell.

I've got the idea for the list reversal and that has lead to the following code:

myreverse list1
 | list1 == []    = list1
 | otherwise      = (myreverse(tail list1)):(head list1)

Unfortunately the above code results in the following error:

Occurs check: cannot construct the infinite type: a = [[a]]
 Expected type: [[a]]
 Inferred type: a
 In the second argument of '(:)', namely '(head list1)'
 In the expression: (myreverse(tail list1)):(head list1)

PS: I get the same sort of error when I run it on a snippet that I wrote called mylast coded below:

mylast list
 | list == []      = []
 | otherwise       = mylast_helper(head list1)(tail list1)

mylast_helper item list2
 | list2 == []     = item
 | otherwise       = mylast_helper(head list2)(tail list2)

Error occurs at the otherwise case of the recursive helper.

EDIT: Thanks for all the input, I guess I forgot to mention that the constraints of the question forbid the use of the ++ operator. I'll keep that in mind for future questions I create.

Cheers, -Zigu

解决方案

You are using the function

(:) :: a -> [a] -> [a]

with ill-typed arguments:

myReverse (tail list1) :: [a]
head list1 :: a

In your function, the list list1 must have type a. Hence the second argument, head list1, must have type [a]. GHC is warning you that it cannot construct the type you have specified for it. The head of a list is structurally smaller than the tail of a list, but you are telling it that the head of a list has type [a], yet the tail of a list has type a.

If you stare closely at your types, however, you will notice that you can append the head of list1 to the recursive call to myreverse using (++):

myReverse xs = case (null xs) of
               True  -> xs
               False -> myReverse (tail xs) ++ [head xs]

Here,

[head xs] :: [a]
myReverse (tail xs) :: [a]

which aligns with the type of append:

(++) :: [a] -> [a] -> [a]

There are much better ways to implement reverse, however. The Prelude defines reverse as a left fold (. Another version of reverse can be implemented using a right fold, and is very similar to your myReverse function:

reverse xs = foldr (\x xs -> xs ++ [x]) [] xs

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