我有以下程序:
{-# LANGUAGE TemplateHaskell #-}
import qualified Data.Map.Strict as Map
import Control.Lens
data MyLabel = MyLabel { _label :: String } deriving (Show, Eq, Ord)
data MyMap = MyMap { _vals :: Map.Map String MyLabel } deriving (Show, Eq, Ord)
makeLenses ''MyLabel
makeLenses ''MyMap
sample :: MyMap
sample = MyMap { _vals = Map.fromList [("foo", MyLabel "bar")] }
现在,我想知道如何f
使用镜头进行转换,例如:
f sample "quux" == MyMap { _vals = Map.fromList [("foo", MyLabel "quux")] }
我了解到at
应该使用Lens库中的函数来修改Maps,因此我正在尝试执行以下操作:
sample ^. vals & at "foo" . label .~ Just "quux"
但这会产生一条错误消息,这对我来说不是很容易理解。什么是正确的方法?
尝试以下尺寸:
{-# LANGUAGE TemplateHaskell #-}
module Main where
import qualified Data.Map.Strict as Map
import Control.Lens
data MyLabel =
MyLabel { _label :: String } deriving (Show, Eq, Ord)
data MyMap =
MyMap { _vals :: Map.Map String MyLabel } deriving (Show, Eq, Ord)
makeLenses ''MyLabel
makeLenses ''MyMap
sample :: MyMap
sample =
MyMap (Map.fromList [("foo", MyLabel "bar")])
main :: IO ()
main =
print (sample & (vals . at "foo" . _Just . label .~ "quux"))
请记住,在设置时,您尝试构建type函数MyMap -> MyMap
。您可以通过将一堆光学元件链接在一起(vals . at "foo" . _Just . label
),然后选择设置器操作(.~
)来实现。您不能像^.
使用setter操作那样混合和匹配getter操作.~
!因此,每个二传手或多或少都是这样的:
foo' = (optic1 . optic2 . optic3 . optic4) .~ value $ foo
-- _________this has type Foo -> Foo___________
为了提高可读性,我们使用&
的翻转版本$
:
foo' = foo & (optic1 . optic2 . optic3 . optic4) .~ value
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我来说两句