Question

我正在编写一个基本的Haskell解释器，并且有以下用例：有两个变量，var1和var2。

if( (typeOf var1 is Integer) and (typeOf var2 is Integer) ) then var1 + var2;

if( (typeOf var1 is String) and (typeOf var2 is String) ) then concatenate var1 to var2;

如何在Haskell中编写它？

代码的一部分：

evaluate:: Expr -> Env -> Val
evaluate expr env = 
  trace("expr= " ++ (show expr) ++ "\n env= " ++ (show env)) $
  case expr of
  Const v -> v
  lhs :+: rhs -> 
    let valLhs = evaluate lhs env
        valRhs = evaluate rhs env
    in case () of
     _ | <both are Integer> ->(IntVal $ (valToInteger valLhs) + (valToInteger valRhs))
       | <both are String>  -> (StringVal $ (valToString valLhs) ++ (valToString valRhs))
       | otherwise....

Answer 1

我没有Val的定义，所以我必须在这里猜测：

case (valLhs, valRhs) of
     (IntVal i1, IntVal i2)       -> IntVal $ i1 + i2
     (StringVal s1, StringVar s2) -> ...

Answer 2

这是使用简单模式匹配评估的混合类型的示例。这不是你问题的确切答案，但也许它可以帮助你。

data Expr a = I Int
            | S String
            | V a
            | Plus (Expr a) (Expr a)
            deriving (Show)

type Env a = a -> Maybe (Expr a)

eval :: Env a -> Expr a -> Expr a
eval _ (I x) = I x
eval _ (S s) = S s
eval _ (Plus (I x) (I y)) = I (x + y)
eval _ (Plus (S x) (S y)) = S (x ++ y)
eval e (Plus (V v) y) = eval e (Plus (eval e (V v)) y)
eval e (Plus x (V v)) = eval e (Plus x (eval e (V v)))
eval _ (Plus _ _) = undefined
eval e (V v) = case e v of Just x -> x
                           Nothing -> undefined

env :: Char -> Maybe (Expr Char)
env 'a' = Just (I 7)
env 'b' = Just (I 5)
env 'c' = Just (S "foo")
env 'd' = Just (S "bar")
env _   = Nothing

检查Haskell脚本中变量的类型

2 个答案: