Haskell 如何在处理递归和类型时减少代码重复
我目前正在为一种编程语言开发一个简单的解释器,我的数据类型如下:Haskell 如何在处理递归和类型时减少代码重复,haskell,functional-programming,dry,code-duplication,recursive-type,Haskell,Functional Programming,Dry,Code Duplication,Recursive Type,我目前正在为一种编程语言开发一个简单的解释器,我的数据类型如下: data Expr = Variable String | Number Int | Add [Expr] | Sub Expr Expr recurseAfter :: (Expr -> Expr) -> Expr -> Expr recurseAfter f x = case f x of Add xs -> Add $ map (recurseAfter f)
data Expr
= Variable String
| Number Int
| Add [Expr]
| Sub Expr Expr
recurseAfter :: (Expr -> Expr) -> Expr -> Expr
recurseAfter f x =
case f x of
Add xs ->
Add $ map (recurseAfter f) xs
Sub x y ->
Sub (recurseAfter f x) (recurseAfter f y)
other -> other
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue =
recurseAfter $ \case
Variable x
| x == name -> Number newValue
other -> other
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd =
recurseAfter $ \case
Sub x (Number y) ->
Add [x, Number (-y)]
other -> other
我有很多函数可以做一些简单的事情,比如:
-- Substitute a value for a variable
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue = go
where
go (Variable x)
| x == name = Number newValue
go (Add xs) =
Add $ map go xs
go (Sub x y) =
Sub (go x) (go y)
go other = other
-- Replace subtraction with a constant with addition by a negative number
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd = go
where
go (Sub x (Number y)) =
Add [go x, Number (-y)]
go (Add xs) =
Add $ map go xs
go (Sub x y) =
Sub (go x) (go y)
go other = other
但是在每一个函数中,我都必须重复递归调用代码的部分,只需对函数的一部分做一点小的更改。有没有更通用的方法?我宁愿不必复制和粘贴此部分:
go (Add xs) =
Add $ map go xs
go (Sub x y) =
Sub (go x) (go y)
go other = other
每次只更改一个案例,因为这样复制代码似乎效率低下
我能想到的唯一解决方案是使用一个函数,该函数首先对整个数据结构调用函数,然后递归调用结果,如下所示:
data Expr
= Variable String
| Number Int
| Add [Expr]
| Sub Expr Expr
recurseAfter :: (Expr -> Expr) -> Expr -> Expr
recurseAfter f x =
case f x of
Add xs ->
Add $ map (recurseAfter f) xs
Sub x y ->
Sub (recurseAfter f x) (recurseAfter f y)
other -> other
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue =
recurseAfter $ \case
Variable x
| x == name -> Number newValue
other -> other
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd =
recurseAfter $ \case
Sub x (Number y) ->
Add [x, Number (-y)]
other -> other
但我觉得应该有一个更简单的方法来做到这一点。我遗漏了什么吗?恭喜你,你刚刚重新发现了变形 这是您的代码,经过重新表述,可以与
递归方案
包一起使用。唉,它并不短,因为我们需要一些样板来让机器工作。(可能有一些自动控制的方法来避免样板文件,例如使用仿制药。我只是不知道。)
下面,您的递归在
被标准的ana
替换之后
我们首先定义递归类型,以及它的不动点函子
{-# LANGUAGE DeriveFunctor, TypeFamilies, LambdaCase #-}
{-# OPTIONS -Wall #-}
module AnaExpr where
import Data.Functor.Foldable
data Expr
= Variable String
| Number Int
| Add [Expr]
| Sub Expr Expr
deriving (Show)
data ExprF a
= VariableF String
| NumberF Int
| AddF [a]
| SubF a a
deriving (Functor)
然后我们用几个实例将两者连接起来,这样我们就可以将Expr
展开成同构的ExprF-Expr
,并将其折叠回去
type instance Base Expr = ExprF
instance Recursive Expr where
project (Variable s) = VariableF s
project (Number i) = NumberF i
project (Add es) = AddF es
project (Sub e1 e2) = SubF e1 e2
instance Corecursive Expr where
embed (VariableF s) = Variable s
embed (NumberF i) = Number i
embed (AddF es) = Add es
embed (SubF e1 e2) = Sub e1 e2
最后,我们修改您的原始代码,并添加两个测试
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue = ana $ \case
Variable x | x == name -> NumberF newValue
other -> project other
testSub :: Expr
testSub = substituteName "x" 42 (Add [Add [Variable "x"], Number 0])
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd = ana $ \case
Sub x (Number y) -> AddF [x, Number (-y)]
other -> project other
testReplace :: Expr
testReplace = replaceSubWithAdd
(Add [Sub (Add [Variable "x", Sub (Variable "y") (Number 34)]) (Number 10), Number 4])
另一种方法是只定义ExprF a
,然后派生type Expr=Fix ExprF
。这节省了上面的一些样板文件(例如两个实例),代价是必须使用Fix(VariableF…
而不是Variable…
,以及其他构造函数的类似代码
使用模式同义词可以进一步缓解这一问题(但要付出更多的代价)
更新:我最终使用模板Haskell找到了automagic工具。这使得整个代码相当简短。请注意,
ExprF
函子和上面的两个实例仍然存在,我们仍然需要使用它们。我们只省去了手动定义它们的麻烦,但这本身就省去了很多工作
{-# LANGUAGE DeriveFunctor, DeriveTraversable, TypeFamilies, LambdaCase, TemplateHaskell #-}
{-# OPTIONS -Wall #-}
module AnaExpr where
import Data.Functor.Foldable
import Data.Functor.Foldable.TH
data Expr
= Variable String
| Number Int
| Add [Expr]
| Sub Expr Expr
deriving (Show)
makeBaseFunctor ''Expr
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue = ana $ \case
Variable x | x == name -> NumberF newValue
other -> project other
testSub :: Expr
testSub = substituteName "x" 42 (Add [Add [Variable "x"], Number 0])
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd = ana $ \case
Sub x (Number y) -> AddF [x, Number (-y)]
other -> project other
testReplace :: Expr
testReplace = replaceSubWithAdd
(Add [Sub (Add [Variable "x", Sub (Variable "y") (Number 34)]) (Number 10), Number 4])
作为替代方法,这也是
uniplate
包的典型用例。它可以使用Data.Data
泛型而不是模板Haskell来生成样板文件,因此如果您为Expr
派生Data
实例:
import Data.Data
data Expr
= Variable String
| Number Int
| Add [Expr]
| Sub Expr Expr
deriving (Show, Data)
import Data.Generics.Uniplate.Data
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue = transform f
where f (Variable x) | x == name = Number newValue
f other = other
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd = transform f
where f (Sub x (Number y)) = Add [x, Number (-y)]
f other = other
然后来自Data.Generics.Uniplate.Data
的transform
函数递归地将函数应用于每个嵌套的Expr
:
import Data.Data
data Expr
= Variable String
| Number Int
| Add [Expr]
| Sub Expr Expr
deriving (Show, Data)
import Data.Generics.Uniplate.Data
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue = transform f
where f (Variable x) | x == name = Number newValue
f other = other
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd = transform f
where f (Sub x (Number y)) = Add [x, Number (-y)]
f other = other
请注意,特别是在replaceSubWithAdd
中,函数f
被编写为执行非递归替换transform
使其在x::Expr
中递归,因此它对helper函数的作用与@chi的答案中的ana
相同:
> substituteName "x" 42 (Add [Add [Variable "x"], Number 0])
Add [Add [Number 42],Number 0]
> replaceSubWithAdd (Add [Sub (Add [Variable "x",
Sub (Variable "y") (Number 34)]) (Number 10), Number 4])
Add [Add [Add [Variable "x",Add [Variable "y",Number (-34)]],Number (-10)],Number 4]
>
这不比@chi的模板Haskell解决方案短。一个潜在优势是uniplate
提供了一些可能有用的附加功能。例如,如果您使用下降
代替变换
,它只变换直接子对象,从而可以控制递归发生的位置,或者您可以使用重写
重新变换变换的结果,直到达到固定点。一个潜在的缺点是“变形”听起来比“单板”冷得多
完整程序:
{-# LANGUAGE DeriveDataTypeable #-}
import Data.Data -- in base
import Data.Generics.Uniplate.Data -- package uniplate
data Expr
= Variable String
| Number Int
| Add [Expr]
| Sub Expr Expr
deriving (Show, Data)
substituteName :: String -> Int -> Expr -> Expr
substituteName name newValue = transform f
where f (Variable x) | x == name = Number newValue
f other = other
replaceSubWithAdd :: Expr -> Expr
replaceSubWithAdd = transform f
where f (Sub x (Number y)) = Add [x, Number (-y)]
f other = other
replaceSubWithAdd1 :: Expr -> Expr
replaceSubWithAdd1 = descend f
where f (Sub x (Number y)) = Add [x, Number (-y)]
f other = other
main = do
print $ substituteName "x" 42 (Add [Add [Variable "x"], Number 0])
print $ replaceSubWithAdd e
print $ replaceSubWithAdd1 e
where e = Add [Sub (Add [Variable "x", Sub (Variable "y") (Number 34)])
(Number 10), Number 4]
制作代码的“提升”版本。在这里,您可以使用决定执行操作的参数(函数)。然后你可以通过将函数传递给提升版本来生成特定的函数。我认为你的语言可以简化。定义
Add::Expr->Expr->Expr
,而不是Add::[Expr]->Expr
,并完全去掉Sub
;虽然这在本例中有效,但我需要能够包含语言其他部分的表达式列表,例如?大多数(如果不是全部的话)链式运算符都可以简化为嵌套的二进制运算符。我认为您的递归后的是ana
伪装的。您可能想看看同构和递归方案
。尽管如此,我认为你的最终解决方案是尽可能短的。切换到官方的递归模式
同构不会节省太多。你真的必须明确定义Expr
,而不是像type Expr=Fix ExprF
这样的东西吗?@chepner我简单地提到了这一点作为替代。对任何事情都必须使用双构造函数有点不方便:Fix
+真正的构造函数。IMO说,将最后一种方法与TH自动化结合使用更好。