Re: [Haskell-cafe] Re: how do you deal with mplus?

So, the law under discussion is (m `mplus` n) >>= f == (m >>= f) `mplus` (n >>= f) Definitely holds for lists and other backtracking searches, but it definitely doesn't hold for non-backtracking monads like Maybe. In particular: divide n 0 = mzero divide n x = return (n `div` x) (Just 0 `mplus` Just 1) >>= divide 100 = Just 0 >>= divide 100 -- definition of MonadPlus Maybe = divide 100 0 -- definition of Monad Maybe = Nothing -- definition of divide whereas (Just 0 >>= divide 100) `mplus` (Just 1 >>= divide 100) = divide 100 0 `mplus` (Just 1 >>= divide 100) = Nothing `mplus` (Just 1 >>= divide 100) = Just 1 >>= divide 100 = divide 100 1 = Just 100 -- ryan