Question

这是关于在Haskell中使用类型类的问题。

我在尝试编译代码时遇到错误（下面）（下面和下面） https://github.com/chrisdew/haskell-sandbox/blob/master/not_working_but_clean.hs ）。

作为刚刚学习Haskell的人，我尝试过遵循GHC的建议，但我认为原因是不同的。

我认为问题在于“IO String”类型或普通类型 'String'可以是'lhello - ＆gt;＆gt;的类型lbracket'，但GHC没有知道哪个。

问题在于无关紧要，任何一种类型都可以正常工作。

我已经发布了代码的工作版本 https://github.com/chrisdew/haskell-sandbox/blob/master/working_but_ugly.hs 。这取代了 - ＆gt;＆gt;之一具有新（非类型类）的运算符运营商' - ＆gt;＆gt;＆gt;'强迫'lhello - ＆gt;＆gt; lbracket'属于'IO 字符串”

我的分析是否正确？或者还有其他事情发生在这里吗？
有没有办法通知GHC'lhello - ＆gt;＆gt;的类型是什么？ lbracket'并不重要，那应该只选择其中之一两种可能性。或者也许是一个LANGUAGE选项我指定'最新声明的匹配的类实例获胜' 如果有什么不确定的话。

谢谢，

克里斯。

错误：

chris@chris-linux-desktop:~/nonworkspace/haskell-sandbox$ ghc
not_working_but_clean.hs

not_working_but_clean.hs:40:16:

   No instance for (Stream (IO String) (IO String) (IO String) d)
     arising from a use of '->>' at not_working_but_clean.hs:40:16-34
   Possible fix:
     add an instance declaration for
     (Stream (IO String) (IO String) (IO String) d)
   In the first argument of '(->>)', namely 'lhello ->> lbracket'
   In the second argument of '($)', namely
       'lhello ->> lbracket ->> putStrLn'
   In a stmt of a 'do' expression:
         forkIO $ lhello ->> lbracket ->> putStrLn

not_working_but_clean.hs:40:16:
   No instance for (Stream d String (IO ()) (IO ()))
     arising from a use of `->>' at not_working_but_clean.hs:40:16-47
   Possible fix:
     add an instance declaration for (Stream d String (IO ()) (IO ()))
   In the second argument of `($)', namely
       `lhello ->> lbracket ->> putStrLn'
   In a stmt of a 'do' expression:
         forkIO $ lhello ->> lbracket ->> putStrLn
   In the expression:
       do { forkIO $ (bracket $ hello) ->> putStrLn;
            forkIO $ lhello ->> lbracket ->> putStrLn;
        forkIO $ bracket hello ->> putStrLn;
            forkIO $ lbracket lhello ->> putStrLn;
          .... }

not_working_but_clean.hs：

{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances,
TypeSynonymInstances, OverlappingInstances #-}
{-# OPTIONS_GHC #-}

module Main (
main
)
where

import Control.Concurrent (forkIO, MVar, newEmptyMVar, putMVar,
takeMVar, ThreadId, threadDelay)
import Control.Monad (forever, liftM)

class Stream a b c d where
   (->>) :: a -> (b -> c) -> d

instance Stream (IO d) d (IO c) (IO c) where
   f ->> g = f >>= g

instance Stream d d (IO c) (IO c) where
   f ->> g = g f

instance Stream d d c c where
   x ->> y = y $ x

-- This simply wraps a string in brackets.
bracket :: String -> String
bracket x = "(" ++ x ++ ")"

lbracket :: IO String -> IO String
lbracket x = liftM bracket x

hello :: String
hello = "Hello World!"

lhello :: IO String
lhello = do return hello

main :: IO ()
main = do
      forkIO $ (bracket $ hello) ->> putStrLn
      forkIO $ lhello ->> lbracket ->> putStrLn
      forkIO $ bracket hello ->> putStrLn
      forkIO $ lbracket lhello ->> putStrLn
      threadDelay 10000000 -- Sleep for at least 10 seconds before exiting.

working_but_ugly.hs：

{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances,
TypeSynonymInstances, OverlappingInstances #-}
{-# OPTIONS_GHC #-}

module Main (
main
)
where

import Control.Concurrent (forkIO, MVar, newEmptyMVar, putMVar,
takeMVar, ThreadId, threadDelay)
import Control.Monad (forever, liftM)

class Stream a b c d where
   (->>) :: a -> (b -> c) -> d

instance Stream (IO d) d (IO c) (IO c) where
   f ->> g = f >>= g

instance Stream d d (IO c) (IO c) where
   f ->> g = g f

instance Stream d d c c where
   x ->> y = y $ x

x ->>> y = y $ x

-- This simply wraps a string in brackets.
bracket :: String -> String
bracket x = "(" ++ x ++ ")"

lbracket :: IO String -> IO String
lbracket x = liftM bracket x

hello :: String
hello = "Hello World!"

lhello :: IO String
lhello = do return hello

main :: IO ()
main = do
      forkIO $ (bracket $ hello) ->> putStrLn
      forkIO $ lhello ->>> lbracket ->> putStrLn
      forkIO $ bracket hello ->> putStrLn
      forkIO $ lbracket lhello ->> putStrLn
      threadDelay 10000000 -- Sleep for at least 10 seconds before exiting.

Answer 1

不，没有办法让GHC翻转硬币并选择一个。

所有实例的类型'c'与类型'd'相同，因此您可以省略类型'd'并在定义Stream时重用类型'c'。

instance Stream d d (IO c) (IO c) where
   f ->> g = g f

instance Stream d d c c where
   x ->> y = y $ x

以上是相同的。 “g f”和“y $ x”是相同的。那么为什么两个不同的实例呢？

在Haskell中输入类 - 当选择的类型无关紧要时，如何让GHC选择类型？

1 个答案: