haskell - 使用免费的 monad 和 GADT 进行 pretty-print

Question

考虑由以下 GADT 定义的表达式仿函数:

{-# LANGUAGE GADTs #-}
{-# LANGUAGE KindSignatures #-}

import Control.Monad.Free

data ExprF :: * -> * where
  Term :: Foo a -> (a -> r) -> ExprF r

instance Functor ExprF where
  fmap f (Term d k) = Term d (f . k)

type Expr = Free ExprF

其中 Foo 定义为

data Foo :: * -> * where
  Bar :: Int    -> Foo Int
  Baz :: Double -> Foo Double

instance Show a => Show (Foo a) where
  show (Bar j) = show j
  show (Baz j) = show j

ExprF 中的 (a -> r) 字段和(否则需要)限制性 GADT 构造函数的组合似乎使得编写 pretty-print 解释器变得不可能:

pretty (Pure r)          = show r
pretty (Free (Term f k)) = "Term " ++ show f ++ pretty (k _)

类型孔是人们所期望的:

Found hole ‘_’ with type: a1
Where: ‘a1’ is a rigid type variable bound by
            a pattern with constructor
              Term :: forall r a. Foo a -> (a -> r) -> ExprF r,
            in an equation for ‘pretty’
            at Test.hs:23:15
Relevant bindings include
  k :: a1 -> Free ExprF a (bound at Test.hs:23:22)
  f :: Foo a1 (bound at Test.hs:23:20)
  pretty :: Free ExprF a -> String (bound at Test.hs:22:1)
In the first argument of ‘k’, namely ‘_’
In the first argument of ‘pretty’, namely ‘(k _)’
In the second argument of ‘(++)’, namely ‘pretty (k _)’

似乎没有办法以其所需的类型为延续赋予一个值。该类型在 f 中编码，我正在使用的其他解释器都以某种方式处理 f 以提取适当类型的值。但是通往 String 表示的路径似乎被阻塞了。

我在这里缺少一些常见的习语吗？如果确实有可能，如何漂亮地打印 Expr 的值？如果不可能，ExprF 的另一种构造可能会捕获相同的结构，但也支持 pretty-print ？

Answer 1

只是在 f 上进行模式匹配。如果这样做，k 的类型将被优化以匹配包含在 Foo 中的类型:

pretty (Pure r)          = show r
pretty (Free (Term f k)) = "Term " ++ show f ++ pretty r where
  r = case f of
    Bar a -> k a
    Baz a -> k a

您可能想分解出这种模式:

applyToFoo :: (a -> r) -> Foo a -> r
applyToFoo f (Bar a) = f a
applyToFoo f (Baz a) = f a

pretty (Pure r)          = show r
pretty (Free (Term f k)) = "Term " ++ show f ++ pretty (applyToFoo k f)