Sengan Baring-Gould writes:
>Is >>= not lazy?
since no experts have answered yet, this newbie will answer.
I think it's strict.
Well, it depends. (>>=) is an overloaded operator, with a different
implementation for every monad -- when you define a monad, you give the
implementation of (>>=). If your implementation is strict (presumably in the
first operand), then (>>=) is strict *at that type*. If your implementation is
lazy, then it isn't. The same goes for (+): at most types (+) is strict, but
if you define your own kind of number with a lazy addition, then on that type
(+) will be lazy.
For many monads, (>>=) *is* strict, which fits with the intuition that it is a
`sequencing' operator. But by no means for all. The simplest counter-example
is the identity monad:
newtype Id a = Id a
instance Monad Id where
return = Id
Id x >>= f = f x
where m>>=f is strict in m only if f is a strict function. A more interesting
example is the state transformer monad:
newtype ST s a = ST (s -> (a,s))
instance Monad (ST s) where
return x = ST (\s -> (x,s))
ST h >>= f = ST (\s -> let (a,s') = h s
ST h' = f a
in h' s')
where once again, the implementation of (>>=) is strict only if f is a
strict function. Hence `lazy state' makes sense!
John Hughes
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