[英]Check the type of a variable in Haskell script
I'm writing a basic Haskell interpreter and there is the following usecase: there are 2 variables, var1 and var2. 我正在写一个基本的Haskell解释器,有以下用例:有2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;
How can I write it in Haskell? 如何在Haskell中编写它?
There's a part of the code: 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....
I don't have the definition of Val
, so I have to guess here: 我没有Val
的定义,所以我不得不在这里猜测:
case (valLhs, valRhs) of
(IntVal i1, IntVal i2) -> IntVal $ i1 + i2
(StringVal s1, StringVar s2) -> ...
Here's an example of mixed types being evaluated using simple pattern-matching. 这是使用简单模式匹配评估混合类型的示例。 It's not an exact answer to your question, but maybe it'll help you. 这不是您问题的确切答案,但也许会对您有所帮助。
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
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