Question

我正在做作业，而且我被困在某事上。

> data Channel a = Chan { cap   :: Int 
>                       , queue :: [a]  }

> data FiniteChan a = IORef (Channel a)

> newFiniteChan :: Int -> IO (FiniteChan a)
> newFiniteChan capacity = do x <- newIORef (Chan {cap = capacity, queue = []})
>                             return x

基本上，我想在调用newFiniteChan时创建对Channel a的引用，我希望FiniteChan a表示该引用（我不能更改newFiniteChan的签名）。当我编译它时，我收到以下错误：

final.lhs:235:38:
    Couldn't match expected type `FiniteChan a'
                with actual type `IORef (Channel a0)'
    In the first argument of `return', namely `x'
    In a stmt of a 'do' block: return x
    In the expression:
      do { x <- newIORef (Chan {cap = capacity, queue = []});
           return x }

我花了很多时间搞清楚问题，但我找不到。我感谢任何帮助。

Answer 1

您必须将构造函数添加到 FiniteChan 数据类型，或将其转换为 type 别名指令，这将编译：

import Data.IORef

data Channel a = Chan { cap   :: Int
                      , queue :: [a]  }

data FiniteChan a = FiniteChan (IORef (Channel a))

newFiniteChan :: Int -> IO (FiniteChan a)
newFiniteChan capacity = do
        x <- newIORef (Chan {cap = capacity, queue = []})
        return $ FiniteChan  x

在创建引用时在Haskell中键入错误

1 个答案: