Created
July 27, 2013 21:17
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Haskell AST basic example
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data Exp r = Const r | Add (Exp r) (Exp r) | Subtract (Exp r) (Exp r) | |
eval :: (Num r) => Exp r -> r | |
eval (Const v) = v | |
eval (Add e1 e2) = (eval e1) + (eval e2) | |
eval (Subtract e1 e2) = (eval e1) - (eval e2) | |
reify :: (Show r) => Exp r -> String | |
reify (Const v) = show v | |
reify (Add e1 e2) = concat ["(",reify e1,"+",reify e2,")"] | |
reify (Subtract e1 e2) = concat ["(",reify e1,"-",reify e2,")"] | |
main :: IO () | |
main = do | |
putStrLn $ reify exp1 | |
putStrLn $ reify exp2 | |
exp1 = Subtract (Add (Const 3) (Const 2)) (Const 1) | |
exp2 = Subtract (Add (Const (3::Double)) (Const 2)) (Const 1) | |
{- | |
Result on GHCi: | |
*Main> main | |
((3+2)-1) | |
((3.0+2.0)-1.0) | |
*Main> :t exp1 | |
exp1 :: Exp Integer | |
*Main> :t exp2 | |
exp2 :: Exp Double | |
*Main> :t (Const 1) | |
(Const 1) :: Num r => Exp r | |
*Main> :t (Const "hello") | |
(Const "hello") :: Exp [Char] | |
*Main> eval (Const "hello") | |
<interactive>:108:1: | |
No instance for (Num [Char]) arising from a use of `eval' | |
Possible fix: add an instance declaration for (Num [Char]) | |
In the expression: eval (Const "hello") | |
In an equation for `it': it = eval (Const "hello") | |
-} |
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