在 Haskell 中反转列表的无限类型错误

Question

我正在尝试实现列表的反转：

myLast :: [a] -> a
myLast [] = error "No end for empty lists!"
myLast [x] = x
myLast (_:xs) = myLast xs

myReverse :: [a] -> [a]
myReverse (x:xs) = myLast xs + myReverse xs

但我收到此错误：

/workspaces/hask_exercises/exercises/src/Lib.hs:42:32: error:
    * Occurs check: cannot construct the infinite type: a ~ [a]
    * In the second argument of `(+)', namely `myReverse xs'
      In the expression: myLast xs + myReverse xs
      In an equation for `myReverse':
          myReverse (x : xs) = myLast xs + myReverse xs
    * Relevant bindings include
        xs :: [a] (bound at src/Lib.hs:42:14)
        x :: a (bound at src/Lib.hs:42:12)
        myReverse :: [a] -> [a] (bound at src/Lib.hs:41:1)
   |
42 | myReverse (x:xs) = myLast xs + myReverse xs
   |                                ^^^^^^^^^^^^

cannot construct the infinite type: a ~ [a] 是什么意思？我经常收到此错误，并想了解它的含义。

Answer 1

(+) :: Num a => a -> a -> a 函数将两个数字（相同类型）相加。所以例如如果a ~ Int，它将添加两个Ints，但不是Int和[Int]。

但是，即使 (+) 运算符例如会将项目添加到列表中，它仍然无法正确反转列表：您的函数没有基本情况对空列表做什么，并且您的递归列表对列表(x:xs) 的第一项x 没有任何作用。

简单的反转方法：

myReverse :: [a] -> [a]
myReverse [] = []
myReverse (x:xs) = myReverse xs ++ [x]

但这并不高效：追加两个项目将花费左列表大小的线性时间。您可以使用累加器：每次进行递归调用时都会更新的参数。这看起来像：

myReverse :: [a] -> [a]
myReverse [] = go []
    where go ys (x:xs) = …
    where go ys [] = …

填写…部分留作练习。

Answer 2

你有

myLast :: [a] -> a 
myReverse :: [a] -> [a]

myReverse (x:xs) = myLast xs    +     myReverse xs
                   \___a___/          \____[a]___/

          (x:xs)  :: [a]
          ---------------
           x      ::  a
             xs   :: [a]                        xs :: [a]               
      myLast      :: [a] -> a         myReverse    :: [a] -> [a]
     -------------------------       ----------------------------
      myLast xs   ::        a         myReverse xs ::        [a]

myReverse (x:xs)                                   ::        [a]

但是

> :t (+)
                               (+) :: Num a =>  a  ->  a  ->  a

表示+左边的东西的类型和右边的东西的类型必须相同。

但它们不可能是：正如我们刚刚在上面看到的，在您的代码中，第一个（myLast xs）是某种类型a，第二个（myReverse xs）是[a] 列表相同的as。

这两个不能相同，因为这意味着

              a  ~  [a]             OK, this is given to us, then
                     a  ~ [a]       we can use it, so that
                 --------------
              a  ~ [[a]]            this must hold;
                     a  ~ [a]       we know this, then
                 --------------
             a  ~ [[[a]]]           this must also hold; and
                     a  ~ [a]       ........
                 --------------     ........
             a ~ [[[[a]]]]          ........
          .........................

以此类推，无限循环，从而使a 成为“无限”类型。因此出现错误。

您可以通过将 + 替换为

来修复它

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

并实施它来做你需要做的事情。

您还需要修复 off-by-one 错误，从而完全忽略接收到的输入中的第一个元素 x。