Re: [Haskell-cafe] Monad Imparative Usage Example

Brian Hulley wrote:

q >>= (\x -> p) For example with the State monad, (q) must be some expression which evaluates to something of the form S fq where fq is a function with type s -> (a,s), and similarly, (\x -> p) must have type a ->S ( s -> (a,s)). If we choose names for these values which describe the types we have: q = S s_as p = a_S_s_as

Sorry I meant: (\x -> p) = a_S_s_as ('p' and 'q' stand for arbitrary expressions that evaluate to monadic values) Regards, Brian. -- Logic empowers us and Love gives us purpose. Yet still phantoms restless for eras long past, congealed in the present in unthought forms, strive mightily unseen to destroy us. http://www.metamilk.com