Question

我有这些函数，stackoverflow中的某个人在尝试解释一个概念时告诉我，但是当我尝试运行它时，我遇到了很多错误。

type World = [String]

removeItem :: String -> World -> IO World
removeItem item world = do
  if elem item world
    then do putStrLn "Item removed."
            return $ delete item world
    else do putStrLn "No such item - " ++ item
            return world

removeTwoItems a b world = do
  world1 <- removeItem a world
  world2 <- removeItem b world
  return world2

look :: World -> IO World
look w = do
  putStrLn $ "You see these items here: " ++ show w
  return w

loop w = do
  words <- fmap words getLine
  case words of
    ("quit":_) -> return ()
    ("look":_) -> do w' <- look; loop w'
    ("remove":item:_) -> do w' <- removeItem item w; loop w'
    _ -> do putStrLn "huh?"; loop w

main = do
  let world0 = ["chair", "bucket"] -- the initial world
  loop world0

错误是：

prac.hs:12:13:
    Couldn't match expected type ‘[Char]’ with actual type ‘IO ()’
    In the first argument of ‘(++)’, namely
      ‘putStrLn "No such item - "’
    In a stmt of a 'do' block: putStrLn "No such item - " ++ item

prac.hs:12:13:
    Couldn't match type ‘[]’ with ‘IO’
    Expected type: IO Char
      Actual type: [Char]
    In a stmt of a 'do' block: putStrLn "No such item - " ++ item
    In the expression:
      do { putStrLn "No such item - " ++ item;
           return world }
    In a stmt of a 'do' block:
      if elem item world then
          do { putStrLn "Item removed.";
               return $ delete item world }
      else
          do { putStrLn "No such item - " ++ item;
               return world }

prac.hs:29:28:
    Couldn't match expected type ‘IO World’
                with actual type ‘World -> IO World’
    Probable cause: ‘look’ is applied to too few arguments
    In a stmt of a 'do' block: w' <- look
    In the expression:
      do { w' <- look;
           loop w' }
Failed, modules loaded: none.

有人可以解释一下这个还是解决这个问题？

Answer 1

函数应用程序的优先级高于任何运算符，因此

putStrLn "No such item - " ++ item

装置

(putStrLn "No such item - ") ++ item

你需要写

putStrLn ("No such item - " ++ item)

或使用$运算符，它是具有较低优先级的函数应用程序相反，像这样：

putStrLn $ "No such item - " ++ item

在函数haskell中获取类型错误

1 个答案: