[英]infinite type error reversing a list in Haskell
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? 我经常收到此错误,并想了解它的含义。
(+):: Num a => a -> a -> a
function 将两个数字(相同类型)加在一起。 因此,例如,如果a ~ Int
,它将两个Int
加在一起,但不是Int
和[Int]
。
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 function 没有基本情况如何处理空列表,并且您的递归列表与第一个无关列表中的项目x
(x:xs)
。
一个简单的反转方法:
myReverse :: [a] -> [a]
myReverse [] = []
myReverse (x:xs) = myReverse xs ++ [x]
但这效率不高:添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。
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
> :t (+)
(+) :: Num a => a -> a -> a
这意味着+
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]
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,无穷无尽,从而使其a
“无限”类型。 因此错误。
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。
您还需要修复一个错误,即完全忽略接收到的输入中的第一个元素x
。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.