CODEWITHSUNDEEP
haskellfunctional-programmingstate
Tri
Tri

Reputation: 113

The >>= implementation of (State s) Monad in Haskell

In the book "Learn You A Haskell for Great Good", The implementation of >>= operator in (State s) Monad is:

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

As we know, the >>= operator has the type >>=::m a -> (a -> m b) -> m b. The second parameter of this operator is a function f whose type is (a -> m b), so why the type of input of function f in the implementation of (State s) is not s -> (a, s) but just the result a?

Thanks for everybody! I think I have figured out how >> = works. We cannot treat the input type of function f as the type contained in Monad, but we should consider it as the input type of monad.

Upvotes: 1

Views: 84

Answers (1)

Daniel Wagner
Daniel Wagner

Reputation: 152737

Since f has type a -> m b (actually, a -> State s b), its input has type a. It is the output whose type has the shape s -> (b, s) (actually, State s b) you mention.

Upvotes: 3

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