Question

是否可以使用匹配/警卫的模式编写函数a？

{-# LANGUAGE PatternGuards #-}
import Control.Monad.State.Strict(State, gets, runStateT)
data MyState = MyState
    { counter :: Int
    } deriving (Show)


a :: State MyState String
a = do
    i <- gets counter
    case i of
        0 -> return "hello"
        1 -> return "bye"

run = runStateT a ( MyState{counter=0} )

我尝试将a写为

a' :: State MyState String
a' | i <- gets counter, i == 0 = return "hello"

但得到了错误：

No instance for (Control.Monad.State.Class.MonadState MyState m0)
  arising from a use of ‘gets’
The type variable ‘m0’ is ambiguous
Note: there are several potential instances:
  instance Control.Monad.State.Class.MonadState s m =>
           Control.Monad.State.Class.MonadState
             s (Control.Monad.Trans.Cont.ContT r m)
    -- Defined in ‘Control.Monad.State.Class’
  instance (Control.Monad.Trans.Error.Error e,
            Control.Monad.State.Class.MonadState s m) =>
           Control.Monad.State.Class.MonadState
             s (Control.Monad.Trans.Error.ErrorT e m)
    -- Defined in ‘Control.Monad.State.Class’
  instance Control.Monad.State.Class.MonadState s m =>
           Control.Monad.State.Class.MonadState
             s (Control.Monad.Trans.Except.ExceptT e m)
    -- Defined in ‘Control.Monad.State.Class’
  ...plus 12 others
In a stmt of a pattern guard for
               an equation for ‘a'’:
  i <- gets counter
In an equation for ‘a'’:
    a' | i <- gets counter, i == 0 = return "hello"

No instance for (Eq (m0 Int)) arising from a use of ‘==’
The type variable ‘m0’ is ambiguous
Relevant bindings include
  i :: m0 Int (bound at src/TestGen/Arbitrary/Helpers/Z.hs:18:6)
Note: there are several potential instances:
  instance Eq a => Eq (GHC.Real.Ratio a) -- Defined in ‘GHC.Real’
  instance (Eq e, Data.Functor.Classes.Eq1 m, Eq a) =>
           Eq (Control.Monad.Trans.Error.ErrorT e m a)
    -- Defined in ‘Control.Monad.Trans.Error’
  ...plus 118 others
In the expression: i == 0
In a stmt of a pattern guard for
               an equation for ‘a'’:
  i == 0
In an equation for ‘a'’:
    a' | i <- gets counter, i == 0 = return "hello"

Answer 1

这是不可能的。模式保护语法中的左箭头大部分与标记中的左箭头无关。

如果您愿意，可以使用新的lambda-case扩展名：

{-# LANGUAGE LambdaCase #-}
a :: State MyState String
a = gets counter >>= \case
        0 -> return "hello"
        1 -> return "bye"

或者多方式，或许？

{-# LANGUAGE MultiWayIf #-}
a :: State MyState String
a = do
    i <- gets counter
    if
      | i == 0 -> return "hello"
      | i == 1 -> return "bye"

Answer 2

没有。这里有一些非常基本的概念不匹配。

模式匹配仅在表达式的最顶部是构造函数函数时才有效，但是do样式块的头部将是普通函数（在本例中是类型类>>=中定义的函数Monad）。

警卫期望值Bool的值，但您要提供的值必须是State MyState Bool类型（因为其中一个独特之处）关于monads的事情是你无法摆脱它们）。所以警卫也永远不会工作。

可以但是可以到达仿函数实例。编曲在前奏中定义; fmap中有一个名为<$>的{{1}}的中缀形式。你可以这样说：

Control.Applicative

或使用a' = process <$> gets counter where process 0 = "hello" process _ = "bye"函数执行任何操作。要获得更像process的内容，您还可以将自己的运算符定义为>>=，然后您可以编写flip fmap。

Answer 3

为什么不写一个帮手？

pureA :: MyState -> String
pureA (MyState 0) = "hello"
pureA (MyState 1) = "bye"
pureA _           = ""

a :: State MyState String
a = fmap doA get

这也遵循将纯逻辑的关注点与不纯的逻辑分开的哲学。

Answer 4

是的，这是可能，但我建议你不要这样做 - 很难跟踪哪一块去哪里。

import Control.Monad.State.Strict(StateT(..))
import Data.Functor.Identity(Identity(..))

data MyState = MyState
    { counter :: Int
    } deriving (Show)

a :: StateT MyState Identity String
a = StateT $ \ s@(MyState i) -> Identity $
  case i of
    0 -> ("hello", s)
    1 -> ("bye", s)

如何在状态monad的字段上进行模式匹配？

4 个答案: