将镜头传递到功能区

时间:2016-03-05 15:36:54

标签: haskell state-monad lens

如何将镜头正确传递到具有状态的功能?让我们考虑下一个代码:

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE FlexibleContexts #-}

import Control.Lens
import Control.Monad.State

data Game = Game { _armies :: [Army]
                 } deriving (Show)

data Army = Army { _troops :: Int
                 } deriving (Show)

makeLenses ''Game
makeLenses ''Army

data BattleResult = Win | Defeat deriving (Show)

offend offender defender = do
  Just ot <- preuse $ offender.troops
  Just dt <- preuse $ defender.troops
  defender.troops.=0 -- doesn't work
  let eval a b
        | a >= b    = return Win
        | otherwise = return Defeat
  eval ot dt

game :: State Game ()
game = do
    armies %= (:) (Army 100)
    armies %= (:) (Army 200)
    q <- offend (armies.ix 0) (armies.ix 1)
    return ()

标记的行会导致下一个错误:

Lens.hs:21:3:
    Couldn't match type ‘Const (Data.Monoid.First Int) s’
                   with ‘Identity s’
    Expected type: (Army -> Const (Data.Monoid.First Int) Army)
                   -> s -> Identity s
      Actual type: (Army -> Const (Data.Monoid.First Int) Army)
                   -> s -> Const (Data.Monoid.First Int) s
    Relevant bindings include
      defender :: (Army -> Const (Data.Monoid.First Int) Army)
                  -> s -> Const (Data.Monoid.First Int) s
        (bound at Lens.hs:18:17)
      offender :: (Army -> Const (Data.Monoid.First Int) Army)
                  -> s -> Const (Data.Monoid.First Int) s
        (bound at Lens.hs:18:8)
      offend :: ((Army -> Const (Data.Monoid.First Int) Army)
                 -> s -> Const (Data.Monoid.First Int) s)
                -> ((Army -> Const (Data.Monoid.First Int) Army)
                    -> s -> Const (Data.Monoid.First Int) s)
                -> m BattleResult
        (bound at Lens.hs:18:1)
    In the first argument of ‘(.)’, namely ‘defender’
    In the first argument of ‘(.=)’, namely ‘defender . troops’

Lens.hs:21:12:
    Couldn't match type ‘Identity Integer’
                   with ‘Const (Data.Monoid.First Int) Int’
    Expected type: (Int -> Identity Integer)
                   -> Army -> Const (Data.Monoid.First Int) Army
      Actual type: (Int -> Const (Data.Monoid.First Int) Int)
                   -> Army -> Const (Data.Monoid.First Int) Army
    In the second argument of ‘(.)’, namely ‘troops’
    In the first argument of ‘(.=)’, namely ‘defender . troops’

如果用armies.ix 0.troops.=0之类的内容替换该行,则通常会编译代码。是否有一些标准工具来解决问题?并且可以在不使用FlexibleContexts的情况下实现相同的算法吗?

1 个答案:

答案 0 :(得分:2)

只需使用类型签名!

此处发生了什么:如果您没有提供签名,GHC将只能推断Rank-1类型。在此示例中,您使用def perm(a): if len(a) <= 1: yield a else: for i in xrange(len(a)): for p in perm(a[:i]+a[i+1:]): yield [a[i]]+p a = [1, 2, 3] for p in perm(a): print(p) [1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1] 作为getter ;因此,编译器推断出defender.troops的getter类型。这是包含defender的错误消息中的丑陋内容。

但是,您还希望将其用作设置器。这只有Const是多态的(因此你可以使用defender仿函数而不是Identity),并且对于多态的参数,你需要Rank-2多态性。

你并不需要担心这种类别理论的魔力,因为镜头库提供了易于使用的同义词。只需写下签名就像始终一样

Const

你得到了正确的多态参数。啊,当然你需要offend :: Traversal' Game Army -> Traversal' Game Army -> State Game BattleResult 扩展名。 -XRankNTypes实际上并不是必需的(但它完全无害,没有理由不使用它)。

如果你问我的话,Hindley-Milner无论如何都是一个奇迹,但它才有用,因为那里有一个明确定义的最普遍的任何表达的签名。这只是Rank-1代码的情况:使用Rank-N,你总是可以投入另一层通用量化。编译器无法知道何时结束此操作!

实际上是一个-XFlexibleContexts,它是一个遍历获取者。 GettingGetter之间的区别在于后者可以是部分的(这是必要的,因为您使用Getting;编译器无法证明索引1处实际存在元素在军队清单中。)