On the other hand, Haskell's "do" notation -- now THAT'S syntactic sugar :-)
-Paul
Ben Lippmeier wrote:
John Goerzen wrote:
Which leaves me with an odd curiosity -- I still can't figure out how state monads are anything but syntactic sugar (and lead more to spaghetti code at that <g>)
Perhaps because state monads = syntactic sugar.
The state monad is just a nice(er) way of passing around some global state (Junk).
Without state monads f :: Junk -> a -> (Junk, b)
With state monads, f :: a -> State Junk b
....
Though if some function doesn't need to 'modify' your Junk, you find yourself having to re-factor things like,
decend :: Junk -> Exp -> Exp decend state (Node a t1 t2) = Node a (decend state t1) (decend state t2)
into
decend :: Exp -> State Junk Exp decend (Node a t1 t2) = do t1' <- decend t1 t2' <- decend t2
return $ Node a t1' t2'
.. which IMHO is not as pretty.
Ben.
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