21 Jan
2012
21 Jan
'12
12:53 p.m.
On 21 Jan 2012, at 21:29, Victor S. Miller wrote:
The "do" notation translates
do {x <- a;f} into
a>>=(\x -> f)
However when we're working in the IO monad the semantics we want requires that the lambda expression be strict in its argument. So is this a special case for IO? If I wanted this behavior in other monads is there a way to specify that?
No, why? The (>>=) combinator (for the IO monad) is strict on it's first argument, that's all. We don't impose any special requirements on the lambda expression.