Haskell - 透鏡，使用'to'功能

Question

我有以下代碼。我希望能夠在給定遊戲狀態時修改主動玩家的生活。我想出了一個activePlayer鏡頭，但當我嘗試，結合了-=操作我收到以下錯誤使用它：

> 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`` 

和有問題的代碼：

{-# 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) 

每個玩家他們依次進行。我需要一次跟蹤所有的球員以及目前處於活躍狀態的球員，這就是我如何構建這一點的原因，儘管我對不同的結構開放，因爲我可能還沒有正確的結構。

Answer 1

當您在鏡頭庫中編寫各種項目時，它們可能會根據某種子類型（參見下文）丟失相應的功能。在這種情況下，你已經由Lens（players）與Getter（to f一些功能f），因此組合僅僅是一個Getter而over作用於鏡頭，既可以獲取和設置。

activePlayer應該形成一個有效的鏡頭，但是，所以你可以把它作爲一個getter/setter手動手動編寫。我在假設索引永遠不會失效的情況下部分寫下它。

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) } 

爲了更好地明白將要發生在lens庫中的子類型，我們可以使用the Big Lattice Diagram from Hackage

每當你將兩個鏡頭類型與您圖表中結束了他們的第一個共同的後代(.) 。所以如果你結合Lens和Prism你可以看到他們的箭頭匯聚在Traversal。如果您合併Lens和Getter（其中to f是），那麼您將獲得Getter，因爲Getter是Lens的直接後代。