功能镜头 [英] Functional lenses

问题描述

有人可以向我解释功能镜头吗？对于谷歌来说这是一个令人惊讶的难题，我还没有取得任何进展。我所知道的是他们提供了类似于OO的get / set功能。

解决方案

镜头由两个功能组成，和setter：

  data Lens ab = Lens {getter :: a  - > b，setter :: b  - > a  - > a} 
  

例如，我们可能会在一对中的第一部分和第二部分使用镜头： p>

  fstLens :: Lens（a，b）a 
 fstLens = Lens fst $ \x（a，b） - > （x，b）
 
 sndLens :: Lens（a，b）b 
 sndLens = Lens snd $ \x（a，b） - > （a，x）
  

镜头真正的便利在于它们组成：

  compose :: Lens bc  - >镜头a b  - >镜头a c 
组合f g = Lens（getter f。getter g）$ 
 \c a  - > setter g（setter fc（getter ga））a 
  

然后它们机械地转换为 State 转换：

  lensGet :: MonadState sm =>镜头s  -   - > m a 
 lensGet =获得。 getter 
 
 lensSet :: MonadState s m =>镜头s b  - > b  - > m（）
 lensSet f =修改。 setter f 
 
 lensMod :: MonadState s m =>镜头s b  - > （b→b）→> m（）
 lensMod f g =修改$ setter f =<< G 。 getter f 
 
（+ =）::（MonadState s m，Num b）=>镜头s b  - > b  - > m（）
 f + = x = lensMod f（+ x）​​
  

Could someone explain functional lenses to me? It's a surprisingly difficult subject to google for and I haven't made any progress. All I know is that they provide similar get/set functionality than in OO.

解决方案

A lens consists of two functions, a getter and a setter:

data Lens a b = Lens { getter :: a -> b, setter :: b -> a -> a }

For example, we might have lenses for the first and second parts of a pair:

fstLens :: Lens (a, b) a
fstLens = Lens fst $ \x (a, b) -> (x, b)

sndLens :: Lens (a, b) b
sndLens = Lens snd $ \x (a, b) -> (a, x)

The real convenience of lenses is that they compose:

compose :: Lens b c -> Lens a b -> Lens a c
compose f g = Lens (getter f . getter g) $
                   \c a -> setter g (setter f c (getter g a)) a

And they mechanically convert to State transitions:

lensGet :: MonadState s m => Lens s a -> m a
lensGet = gets . getter

lensSet :: MonadState s m => Lens s b -> b -> m ()
lensSet f = modify . setter f

lensMod :: MonadState s m => Lens s b -> (b -> b) -> m ()
lensMod f g = modify $ setter f =<< g . getter f

(+=) :: (MonadState s m, Num b) => Lens s b -> b -> m ()
f += x = lensMod f (+ x)

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