On 5/4/08, Iavor Diatchki <[EMAIL PROTECTED]> wrote: > >From the monad law we can conclude only that "(>>= return)" is strict, > not (>>=) in general. > For example, (>>=) for the reader monad is not strict in its first argument: > > m >>= f = \r -> f (m r) r > > So, "(undefined >> return 2) = (return 2)"
That's not even true here, though, with regards to seq. undefined >>= return = \r -> return (undefined r) r = \r -> const (undefined r) r = \r -> undefined r But seq undefined 0 = _|_ seq (undefined >>= return) 0 = seq (\r -> undefined r) 0 = 0 The monad laws just aren't true for many monads once seq is a possibility. -- ryan _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
