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我有以下代码。我希望能够在给定游戏状态时修改活跃玩家的生活。我想出了一个activePlayer镜头，但是当我尝试将它与-=操作员结合使用时，我收到以下错误：

> over (activePlayer.life) (+1) initialState 
<interactive>:2:7:
    No instance for (Contravariant Mutator)
      arising from a use of `activePlayer'
    Possible fix:
      add an instance declaration for (Contravariant Mutator)
    In the first argument of `(.)', namely `activePlayer'
    In the first argument of `over', namely `(activePlayer . life)'
    In the expression: over (activePlayer . life) (+ 1) initialState``

和有问题的代码：

{-# LANGUAGE TemplateHaskell #-}
module Scratch where

import Control.Lens
import Control.Monad.Trans.Class
import Control.Monad.Trans.State
import Data.Sequence (Seq)
import qualified Data.Sequence as S

data Game = Game
    { _players :: (Int, Seq Player) -- active player, list of players
    , _winners :: Seq Player
    }
    deriving (Show)

initialState = Game
    { _players = (0, S.fromList [player1, player2])
    , _winners = S.empty
    }

data Player = Player
    { _life :: Integer
    }
    deriving (Show, Eq)

player1 = Player
    { _life = 10
    }

player2 = Player
    { _life = 10
    }

makeLenses ''Game
makeLenses ''Player

activePlayer
  :: (Functor f, Contravariant f) =>
       (Player -> f Player) -> Game -> f Game
activePlayer = players.to (\(i, ps) -> S.index ps i)

每个玩家按顺序轮到他们。我需要同时跟踪所有玩家以及当前活跃的玩家，这就是我如何构建它的原因，尽管我对不同的结构持开放态度，因为我可能还没有合适的结构。

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1 回答 1

12

当您在镜头库中组合各种项目时，(.)它们可能会根据一种子类型而失去功能（见下文）。在这种情况下，您已经将Lens( players) 与Getter(to f用于某些功能f) 组合在一起，因此组合只是Getter一会儿over作用于可以获取和设置的镜头。

activePlayer但是，应该形成一个有效的镜头，因此您可以手动将其编写为 getter/setter 对。我在假设索引永远不会无效的情况下将其部分写在下面。

activePlayer :: Lens' Game Player
activePlayer = lens get set 
  where
    get :: Game -> Player
    get (Game { _players = (index, seq) }) = Seq.index seq index

    set :: Game -> Player -> Game
    set g@(Game { _players = (index, seq) }) player = 
      g { _players = (index, Seq.update index player seq) }

为了更好地理解lens库中出现的子类型，我们可以使用Hackage 中的 Big Lattice Diagram

Hackage 的大格子图

每当您将两种镜头类型组合在一起时，(.)您最终都会在该图表中获得它们的第一个共同后代。所以如果你结合起来Lens，Prism你可以看到它们的箭头会聚在 上Traversal。如果你结合Lensand Getter(of which to fis) 那么你会得到一个GettersinceGetter是Lens.

于 2013-12-11T20:20:02.550 回答