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

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?

That definition of `pop` terrifies me; its a deeply hidden partial function. I know it wouldn't be any better `fail`ing in `Identity` for a stack underflow, but it would be less terrifying. — Cirdec, Sep 08 '15 at 21:35

score 2 · Accepted Answer · answered Sep 08 '15 at 21:25

2

The compiler can figure out the type to fill in a blank. If I add a hole _ to the code

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

The compiler tells me it should have the type Integer.

   Found hole `_' with type: Integer

What Integer could you put there that would be the sum of the values popped off the stack?

answered Sep 08 '15 at 21:25

Cirdec

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Why do you think `s1` and `s2` aren't `Integer`s? – Cirdec Sep 08 '15 at 21:37
How can i get integer value from the type State in this place? – spiridoncha Sep 08 '15 at 21:40
`pop` is a `State Stack Integer`. When you `>>=` (pronounced bind) it to a function `\x -> ...` the `x` is an `Integer`. The `...` needs to be a `State Stack a` for some type `a`. – Cirdec Sep 08 '15 at 21:44
Thank you!! yes, of course s1 and s2 are integer. i'm just stupid :\ – spiridoncha Sep 08 '15 at 21:48

score 0 · Answer 2 · answered Sep 08 '15 at 22:16

You mean this?

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]

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

2 Answers2