打印免费单片机

Question

可以将免费单片机翻译为任何其他单片机，但给定类型为Free f x的值时，我想打印整棵树，而不是将生成的AST的每个节点映射到另一单片机中的某个其他节点。

加布里埃尔冈萨雷斯uses值直接

showProgram :: (Show a, Show r) => Free (Toy a) r -> String 
showProgram (Free (Output a x)) = 
    "output " ++ show a ++ "\n" ++ showProgram x 
showProgram (Free (Bell x)) = 
    "bell\n" ++ showProgram x 
showProgram (Free Done) = 
    "done\n" 
showProgram (Pure r) = 
    "return " ++ show r ++ "\n" 

可以抽象出来作为

showF :: (x -> b) -> ((Free f x -> b) -> f (Free f x) -> b) -> Free f x -> b 
showF backLiftValue backLiftF = fix (showFU backLiftValue backLiftF) 
    where 
     showFU :: (x -> b) -> ((Free f x -> b) -> f (Free f x) -> b) -> (Free f x -> b) -> Free f x -> b 
     showFU backLiftValue backLiftF next = go . runIdentity . runFreeT where 
      go (FreeF c) = backLiftF next c 
      go (Pure x) = backLiftValue x 

这是容易的，如果我们有像（使用Choice x = Choice x x作为仿）

一个多态函数调用

showChoice :: forall x. (x -> String) -> Choice x -> String 
showChoice show (Choice a b) = "Choice (" ++ show a ++ "," ++ show b ++ ")" 

但是，这似乎相当复杂用于简单操作... 还有什么其他方法可以从f x -> b到Free f x -> b？

Answer 1

使用iter和fmap：

{-# LANGUAGE DeriveFunctor #-} 

import Control.Monad.Free 

data Choice x = Choice x x deriving (Functor) 

-- iter :: Functor f => (f a -> a) -> Free f a -> a 
-- iter _ (Pure a) = a 
-- iter phi (Free m) = phi (iter phi <$> m) 

showFreeChoice :: Show a => Free Choice a -> String 
showFreeChoice = 
     iter (\(Choice l r) -> "(Choice " ++ l ++ " " ++ r ++ ")") 
    . fmap (\a -> "(Pure " ++ show a ++ ")") 

fmap转换从Free f a到Free f b，并iter没有休息。你可以考虑这一点，也许会得到更好的表现：

iter' :: Functor f => (f b -> b) -> (a -> b) -> Free f a -> b 
iter' f g = go where 
    go (Pure a) = g a 
    go (Free fa) = f (go <$> fa)