Ok I got the confusion clear, I was referring to the wrong definitions of push and pop

pop = Stack $ (\x:xs) -> (x,xs)
push a = State $ \xs -> ((), xs)

Now the do operation makes sense coz now I can relate all this to simple monads….

(>>=) :: Monad m => m a -> (a -> m b) -> m b
Where m = state s being shoved around!

-Animesh


On Mar 8, 2015, at 1:34 AM, Animesh Saxena <animeshsaxena@icloud.com> wrote:

Thanks for the links, I am able to get my head around this a bit. 

Here's my simple question
1. In the stack example what is the state. As per my understanding the state is the new stack.
If above statement is correct than by definition of >>=

  (>>=) :: State s a -> (a -> State s b) -> State s b

push 3 stack 
This returns ((), newstack1)
Now in this case the state or context is newstack1 which has to be passed to the next function if i apply >>=.
push 3 stack >>= pop newstack1
This might make sense coz I am shoving (or binding) the state to the pop function. State in this case is the stack which is being passed around or shoved around. Problem is pop doesn't need "a" argument like the definition of (>>=) indicates. It can very well produce a new state or return  a new state. 

So if state = stack then this makes sense to me, where am i wrong...??

push 3 stack >>= pop newstack1

On Mar 06, 2015, at 09:09 PM, "Sumit Sahrawat, Maths & Computing, IIT (BHU)" <sumit.sahrawat.apm13@iitbhu.ac.in> wrote:

I won't comment on what state exactly is, but you can read up on that and gain some intuition here: https://en.wikibooks.org/wiki/Haskell/Understanding_monads/State
It's helpful to implement it using a pen and paper, and consider how the state flows and gets transformed.

According to the below example,

  stackManip stack =
    let ((), newStack1) = push 3 stack
        (a, newStack2)  = pop newStack1
    in pop newStack2

We get,

  push :: a -> Stack a -> ((), Stack a)    -- Assuming 'Stack a' is a defined datatype
  pop  :: Stack a -> (a, Stack a)          -- Representing a stack with elements of type 'a'

Thus,

    push 3 >>= pop
~~  (Stack a -> ((), Stack a)) >>= (Stack a -> (a, Stack a))               { Replacing by types }

Bind (>>=) has the type, (for "State s")

  (>>=) :: State s a -> (a -> State s b) -> State s b

This is a type mismatch. The conversion to do syntax is at fault here.
First, you must write the computation using bind (>>=), and then convert to do-notation.

On 7 March 2015 at 10:12, Animesh Saxena <animeshsaxena@icloud.com> wrote:
I am trying to relate the state monad to a stack example and somehow found it easy to get recursively confused!

instance Monad (State s) where
    return x = State $ \s -> (x,s)
    (State h) >>= f = State $ \s -> let (a, newState) = h s
                                                    (State g) = f a
                                                     in g newState

Considering the stack computation

stackManip stack = let 
              ((), newStack1) = push 3 stack
              (a, newStack2) = pop newStack1
               in pop newStack2

in do notation this would become 
do  
   push 3  
   a <- pop  
   pop

If I consider the first computation push 3 >>= pop and try to translate it to the definition there are problems....
Copy paste again, I have 
    (State h) >>= f = State $ \s -> let (a, newState) = h s
                                                    (State g) = f a
                                                     in g newState

f is the push function to which we are shoving the old state. I can't exactly get around to what exactly is the state computation h? Ok assuming it's something which gives me a new state, but then thats push which is function f. 
Then push is applied to a which is assuming 3 in this case. This gives me a new state, which I would say newStack1 from the stockManip above.

Then somehow I apply g to newState?? All the more confusion. Back to the question what exactly is state computation and how is it different from f? It seems to be the same function?


-Animesh




                           

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--
Regards

Sumit Sahrawat
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