Why use such a peculiar function type in monads?

Question

New to Haskell, and am trying to figure out this Monad thing. The monadic bind operator -- >>= -- has a very peculiar type signature:

(>         


        

Answer 1

It's much more symmetric if you think in terms the following derived function (from Control.Monad):

(>=>) :: Monad m => (a -> m b) -> (b -> m c) -> (a -> m c)
(f >=> g) x = f x >>= g

The reason this function is significant is that it obeys three useful equations:

-- Associativity
(f >=> g) >=> h = f >=> (g >=> h)

-- Left identity
return >=> f = f

-- Right identity
f >=> return = f

These are category laws and if you translate them to use (>>=) instead of (>=>), you get the three monad laws:

(m >>= g) >>= h = m >>= \x -> (g x >>= h)

return x >>= f = f x

m >>= return = m

So it's really not (>>=) that is the elegant operator but rather (>=>) is the symmetric operator you are looking for. However, the reason we usually think in terms of (>>=) is because that is what do notation desugars to.