用延续重写代码

Question

我有一些评估原始程序的代码。程序是一个语句列表（表达式、块、返回语句）。评估结果是最后评估的表达式。评估者还应正确对待return 语句（即在第一次出现return 后停止评估）。

为了实现这个逻辑，我传递了特殊的回调函数 (NextStep)，它在当前语句之后进行下一步评估。处理return语句时不调用下一步：

data Statement = 
      Expr Int
    | Block [Statement]
    | Return Int
    deriving (Show, Eq)

data Value = 
      Undefined
    | Value Int
    deriving (Show, Eq)

type NextStep = Value -> Value

evalStmt :: Statement -> NextStep -> Value
evalStmt (Expr val) next = 
    let res = Value val
    in next res
evalStmt (Block stmts) next = evalBlock stmts next
evalStmt (Return val) next = Value val

evalBlock :: [Statement] -> NextStep -> Value
evalBlock [] next = next Undefined
evalBlock [st] next = evalStmt st next
evalBlock (st:rest) next = evalStmt st $ \ _ -> evalBlock rest next

evalProgram stmts = evalBlock stmts id

prog1 = [Expr 1, Block [Return 3, Expr 2], Expr 4] 
evalProg1 = evalProgram prog1 -- result will be Value 3

问题是如何用 continuation monad 重写这段代码？我想摆脱在evalStmt 和evalBlock 函数中显式传递的NextStep 回调。有可能吗？

Answer 1

翻译相当机械。

请记住，在延续单子中，return 将值提供给延续。

evalStmt :: Statement -> Cont Value Value
evalStmt (Expr val) = 
    let res = Value val
    in return res
evalStmt (Block stmts) = evalBlock stmts
evalStmt (Return val) = cont $ \_ -> Value val

evalBlock :: [Statement] -> Cont Value Value
evalBlock [] = return Undefined
evalBlock [st] = evalStmt st
evalBlock (st:rest) = evalStmt st >> evalBlock rest

evalProgram :: [Statement] -> Value
evalProgram stmts = runCont (evalBlock stmts) id

为了模拟提前返回，我们只是忽略给Return val 的延续，只返回我们拥有的值。