how to increment a variable in functional programming

As a general rule, you don't (and you don't need to). However, in the interests of completeness.

import Data.IORef
main = do
    i <- newIORef 0       -- new IORef i
    modifyIORef i (+1)    -- increase it by 1
    readIORef i >>= print -- print it

However, any answer that says you need to use something like MVar, IORef, STRef etc. is wrong. There is a purely functional way to do this, which in this small rapidly written example doesn't really look very nice.

import Control.Monad.State
type Lens a b = ((a -> b -> a), (a -> b))
setL = fst
getL = snd
modifyL :: Lens a b -> a -> (b -> b) -> a
modifyL lens x f = setL lens x (f (getL lens x))
lensComp :: Lens b c -> Lens a b -> Lens a c
lensComp (set1, get1) (set2, get2) =         -- Compose two lenses
    (\s x -> set2 s (set1 (get2 s) x)        -- Not needed here
     , get1 . get2)                          -- But added for completeness

(+=) :: (Num b) => Lens a b -> Lens a b -> State a ()
x += y = do
    s <- get
    put (modifyL x s (+ (getL y s)))

swap :: Lens a b -> Lens a b -> State a ()
swap x y = do
    s <- get
    let x' = getL x s
    let y' = getL y s
    put (setL y (setL x s y') x')

nFibs :: Int -> Int
nFibs n = evalState (nFibs_ n) (0,1)

nFibs_ :: Int -> State (Int,Int) Int
nFibs_ 0 = fmap snd get -- The second Int is our result
nFibs_ n = do
    x += y       -- Add y to x
    swap x y     -- Swap them
    nFibs_ (n-1) -- Repeat
  where x = ((\(x,y) x' -> (x', y)), fst)
        y = ((\(x,y) y' -> (x, y')), snd)

Simple way is to introduce shadowing of a variable name:

main :: IO ()                  -- another way, simpler, specific to monads:
main = do                         main = do
    let i = 0                         let i = 0
    let j = i                         i <- return (i+1)
    let i = j+1                       print i
    print i                    -- because monadic bind is non-recursive

Prints 1.

Just writing let i = i+1 doesn't work because let in Haskell makes recursive definitions — it is actually Scheme's letrec. The i in the right-hand side of let i = i+1 refers to the i in its left hand side — not to the upper level i as might be intended. So we break that equation up by introducing another variable, j.

Another, simpler way is to use monadic bind, <- in the do-notation. This is possible because monadic bind is not recursive.

In both cases we introduce new variable under the same name, thus "shadowing" the old entity, i.e. making it no longer accessible.

How to "think functional"

One thing to understand here is that functional programming with pure — immutable — values (like we have in Haskell) forces us to make time explicit in our code.

In imperative setting time is implicit. We "change" our vars — but any change is sequential. We can never change what that var was a moment ago — only what it will be from now on.

In pure functional programming this is just made explicit. One of the simplest forms this can take is with using lists of values as records of sequential change in imperative programming. Even simpler is to use different variables altogether to represent different values of an entity at different points in time (cf. single assignment and static single assignment form, or SSA).

So instead of "changing" something that can't really be changed anyway, we make an augmented copy of it, and pass that around, using it in place of the old thing.