Hello,
On 4/15/07, David Powers
so... this is likely a question based on serious misunderstandings, but can anyone help me understand the exact mechanism by which monads enforce sequencing?
Monads do not enforce sequencing. In general, data dependencies enforce sequencing (i.e. if expression x depends upon expression y then expression y will have to be evaluated first). Consider: let x = case y of Just y' -> f y' Nothing -> g y = some code in more stuff Here y must be evaluated before x because x needs to look at y in order to compute.
Specifically, I'm confused by the >> operator. If I understand things properly f a >> g expands to something like:
f >>= \_ -> g
What I'm missing is how the expansion of f is ever forced under lazy evaluation. Since the result is never used, doesn't it just stay as a completely unevaluated thunk?
(>>=) is an overloaded function. Some instances of it will cause f to be evaluated, others won't. Consider the State monad: instance Monad (State s) where return a = State $ \s -> (a, s) m >>= k = State $ \s -> case runState m s of (a, s') -> runState (k a) s' Note that (>>=) causes m to be evaluated (up to a pair) before evaluating k because (>>=) needs to look at the result of m. An example of a monad in which (>>=) doesn't force evaluation of the first argument before the second is the Identity monad: instance Monad Identity where return a = Identity a m >>= k = k (runIdentity m) Note that (>>=) actually forces the evaluation of k before m. -Jeff