如何解决在 Haskell 中反转列表的无限类型错误
我正在尝试实现列表的反向:
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?我经常收到这个错误,想了解它的含义。
解决方法
(+) :: Num a => a -> a -> a
函数将两个数字(相同类型)相加。例如,如果 a ~ Int
,它会将两个 Int
相加,但不会将 Int
和 [Int]
相加。
但是,即使 (+)
运算符例如将一个项目添加到列表中,它仍然无法正确反转列表:您的函数没有基本情况要为一个列表做什么空列表,并且您的递归列表对列表 x
的第一项 (x:xs)
不执行任何操作。
简单的反转方法:
myReverse :: [a] -> [a]
myReverse [] = []
myReverse (x:xs) = myReverse xs ++ [x]
但这效率不高:附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 …
部分的地方留作练习。
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+
左边的事物的类型和右边的事物的类型必须相同。
但它们不能是:正如我们刚刚在上面看到的,在您的代码中,第一个(myLast xs
的)是某种类型的 a
,第二个(myReverse xs
的)是[a]
那些相同 a
的列表。
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,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
(+++) :: a -> [a] -> [a]
并实施它来做你需要它做的事情。
您还需要修复您的逐个错误,从而完全忽略接收到的输入中的第一个元素 x
。
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