I gave an expansion of the state monad for a different computation on Reddit some time ago. Perhaps it will be useful to you: http://www.reddit.com/r/haskell/comments/25fnrj/nicta_course_help_with_state... Best, Ben On Sat, 7 Mar 2015 5:09 am 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 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 _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners