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
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
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
-- Regards Sumit Sahrawat