State in Haskell is simple

state-is-simple.hs

#!/usr/bin/env stack
-- stack --resolver lts-12.24 script

-- If stack isn't installed on your system and you can't/won't get it, install
-- a reasonably-modern GHC and use this instead of the above two lines:
-- #! /usr/bin/env runhaskell

import Control.Monad.State


main :: IO ()
main = do
  {-
    We've seen examples of how to build up the State monad by hand which is
    helpful to understand what's happening under the State monad covers.
    I wanted to illustrate just how minimal and simple it is to use the state
    monad in real world code.
  -}

  let startingState = 0
  let result = runState ( do
      addFive
      multThree
      ) startingState

  -- This code is operating entirely on the state, the return value, in the
  -- left side of the tuple, is in this case a useless () value that we'll drop
  print . snd $ result

  {-
    One case for using the State monad is that you're going to need the state
    in some functions that this computation will call (signified here by
    addFive and multThree) and passing it around explicitly will be ugly.
    There are other reasons you may want State monad specifically but it's not
    for everything.
  -}


  -- Sometimes a stateful task is too trivial for even that much code.

  -- In simple cases, function composition will suffice
  print $ (* 3) . (+ 5) $ 0

  -- Another example uses a right fold and the function application function to
  -- thread the state through a list of functions.
  print $ foldr ($) 0 [(* 3), (+ 5)]


  -- All three of these examples output the same value


-- Functions that need the state above, no need to pass explicitly
addFive :: State Int ()
addFive = do
  n <- get
  put $ n + 5

multThree :: State Int ()
multThree = modify (* 3)