如何解决这些 Haskell 类型错误
所以我一直在摆弄 Haskell，并且在我的代码中遇到了这个奇怪的错误。
““IO”未应用于足够的类型参数
应为“?”类型，但“IO”类型为“->”
在'loop'的类型签名中：loop :: State -> IO”
这是代码
import System.IO
data State = State [Int] Int Int deriving (Show)
main = do
loop (State [] 0 0)
loop::State -> IO
loop state = do
putStr "file: "
f <- getLine
handle <- openFile f ReadMode
cde <- hGetContents handle
hClose handle
putStrLn cde
loop state
我该如何修复此错误？此外，任何有关类型的见解将不胜感激。
So I've been messing around with Haskell, and I've come across this strange error in my code.
" 'IO' is not applied to enough type arguments
Expected kind '?', but 'IO' has kind '->'
In the type signature for 'loop': loop :: State -> IO"
Here is the Code
import System.IO
data State = State [Int] Int Int deriving (Show)
main = do
loop (State [] 0 0)
loop::State -> IO
loop state = do
putStr "file: "
f <- getLine
handle <- openFile f ReadMode
cde <- hGetContents handle
hClose handle
putStrLn cde
loop state
How do I fix this error? Also, any insight on kinds would be greatly appreciated.
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IO是一个类型构造函数，这意味着它需要一个参数才能成为一种类型。所以：是类型，但 IO 本身不是类型。
IO的种类是* -> *，这就像说它是一个接受类型并返回类型的函数。我建议更改
为
(
()是“单位类型”，它只有一个值（也称为()），通常在void 的情况下使用将用于类 C 语言）IOis a type constructor, which means that it needs an argument in order to become a type. So:are types, but
IOby itself is not.The kind of
IOis* -> *, which is like saying it is a function that takes a type and returns a type.I would suggest changing
to
(
()is the "unit type", it has only one value (also called()), and is typically used wherevoidwould be used in C-like languages)IO是一个 类型构造函数，而不是完整类型。您应该声明其中
()是单元类型；仅具有一个值的类型，也拼写为()。这是永恒循环或任何其他不返回（有意义的）值的函数的适当类型。IOis a type constructor, not a full type. You should declarewhere
()is the unit type; the type with only one value, also spelled(). That's the appropriate type for an eternal loop or any other function that does not return a (meaningful) value.正如其他人提到的，IO 是类型构造函数，而不是类型。所以你必须将它应用到其他类型。
IO Foo类型的值意味着它是一个可能执行某些 I/O，然后返回Foo类型的值的计算。luqui 和 larsman 建议您应该使用
()作为返回值。我认为以下类型是永远循环的函数的更好替代方案：请注意，该函数现在的返回值是多态的。这种类型比返回
()提供更多信息。为什么？因为这种类型的函数必须是一个循环函数。无法用这种类型实现终止函数。该函数的用户将从类型中立即看出它是一个循环函数。因此，您可以免费获得一些此类类型的文档。As others have mentioned,
IOis a type constructor, not a type. So you have to apply it to some other type. A value of typeIO Foomeans that it is a computation which potentially does some I/O and then returns a value of typeFoo.luqui and larsman suggested that you should use
()as a return value. I think the following type is a better alternative for a function that loops forever:Note that the function now is polymorphic in the return value. This type is much more informative than having it return
(). Why? Because a function of this type must be a looping function. There is no way to implement a terminating function with this type. A user of this function will see immediately from the type that it is a looping function. So you get some documentation for free with this type.