我如何从堆栈中弹出两个元素，然后将它们作为值的总和推回堆栈？

Question

import Control.Monad.State

type Stack = [Integer]

pop :: State Stack Integer
pop = state $ \(x:xs) -> (x, xs)

push :: Integer -> State Stack ()
push x = state $ \xs -> ((), (x:xs))

main :: IO()
main = print $ runState `enter code here` [1,2,3,4]

using "pop >>= (\s1 -> pop >>= (\s2 -> push enter code here)" what should i write here?

Answer 1

编译器可以figure out the type to fill in a blank。如果我在代码

中添加一个洞_

add = pop >>= \s1 -> pop >>= \s2 -> push _

编译器告诉我它的类型应该是 Integer.

   Found hole `_' with type: Integer

你能把什么 Integer 放在那里作为从堆栈弹出的值的总和？

Answer 2

你是这个意思？

popAndSum :: Int -> [Int] -> [Int]
popAndSum elem list = fst xs ++ [sum']
        where
            xs = splitAt ((length list) - elem) list
            sum' = foldl (+) 0 $ snd xs

popAndSum 2 [1,2,3,4,5] = [1,2,3,9]

popAndSum 3 [1,2,3,4,5] = [1,2,12]

我如何从堆栈中弹出两个元素，然后将它们作为值的总和推回堆栈？

how can i pop two elements from a stack and then push in stack them back as sum of values?

haskell

state-monad