역 폴란드 식 의 실현

data Lex = Number Double Lex
 | Plus Lex
 | Times Lex
 | End deriving Show

lexRPN :: String -> Lex
lexRPN = go . words
 where 
  go ("*":rest) = Times (go rest)
  go ("+":rest) = Plus (go rest)
  go (num:rest) = Number (read num) (go rest)
  go [] = End

evalRPN :: Lex -> Double
evalRPN = go [] where
  go stack (Number num rest) = go (num : stack) rest
  go (o1:o2:stack) (Plus rest)= let r = o1 + o2 in r `seq` go (r : stack) rest
  go (o1:o2:stack) (Times rest)= let r = o1 * o2 in r `seq` go (r : stack) rest
  go [res] End = res 

ghci> lexRPN "3 1 2 + *"
Number 3.0 (Number 1.0 (Number 2.0 (Plus (Times End))))
ghci> evalRPN $ lexRPN "3 1 2 + *"
9.0
ghci> evalRPN $ lexRPN "5 6 2 + 7 * *"
280.0
ghci>