Haskell-计算字符串“ let X = 3 in + XX”

Question

嗨，我需要帮助来计算类型。 假设整个程序采用一个字符串，将其解析，最后计算一个值。 我开始的字符串可能是这样的：“ let X = + 1 2 in * X 2 - X ”。 当我解析它时，我会得到：“ Let (Vari X) (Sum (Lit 1) (Lit 2)) (Mul (Vari X) (Lit 2))) ”。 目前，我可以解析“ * + 2 3 * 2 + 6-2”和之前的表达式。 但是我无法计算前面的表达式“ let X = + 1 2 in * X 2 - X ”。 如果有人能指出正确的方向，我会很高兴，因为现在，我真的不知道如何进行这项工作。 谢谢

码：

data Chara = A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z deriving (Eq, Show)

data Number = Single Int | Many Int Number deriving (Eq, Show)

data Expr = Lit Int | Sub Expr | Sum Expr Expr | Mul Expr Expr | Vari Chara | Let Expr Expr Expr 
    deriving Show


--Want to calculate this
--Let (Vari X) (Sum (Lit 1) (Lit 2)) (Mul (Vari X) (Lit 2)))


calculate :: Expr -> Int 
calculate (Sub a) = let e = calculate a in (-e)
calculate (Sum a b) = let e = calculate a 
                          r = calculate b 
                      in (e+r)          
calculate (Mul a b) = let e = calculate a 
                          r = calculate b 
                      in (e*r)
calculate (Lit a) = a       

Answer 1

您需要在AST中执行变量替换 。 即你需要一个功能

substitute :: (Chara, Expr) -> Expr -> Expr

给定一对要替换的变量和用其 替换的 表达式，它们将遍历树并执行该替换。 这基本上意味着：如果找到Vari ，只需将其替换为替换项即可。 如果找到带有子表达式之类的东西，例如Sum ab ，则递归到这些子表达式中，然后用结果重建运算符，即

        Sum (substitute s a) (substitute s b)

然后， calculate (Let var subst expr) = ...是对substitute函数的直接调用。

Answer 2

奇迹般有效

calculate :: Expr -> Int 
calculate (Sub a) = let e = calculate a in (-e)
calculate (Sum a b) = let e = calculate a 
                          r = calculate b 
                      in (e+r)          
calculate (Mul a b) = let e = calculate a 
                          r = calculate b 
                      in (e*r)
calculate (Let a b c) = calculate (substitute (getCharaFromExpr a) b c)
calculate (Lit a) = a       


substitute :: Chara -> Expr -> Expr -> Expr 
substitute x y (Lit a) = Lit a
substitute x y (Vari a) | x == a = y
                        | otherwise = Vari a 
substitute x y (Sum a b) = Sum (substitute x y a) (substitute x y b)
substitute x y (Mul a b) = Mul (substitute x y a) (substitute x y b)
substitute x y (Sub a) = Sub (substitute x y a)
substitute x y (Let a b c) = Let (substitute x y a) (substitute x y b) (substitute x y c)