On Sun, 2009-03-15 at 21:43 +0100, Daniel Fischer wrote: > Am Sonntag, 15. März 2009 21:25 schrieb Jonathan Cast: > > On Sun, 2009-03-15 at 13:02 -0700, Ryan Ingram wrote: > > > > > Furthermore, due to the monad laws, if f is total, then reordering the > > > (x <- ...) and (y <- ...) parts of the program should have no effect. > > > But if you switch them, the program will *always* print 0. > > > > I'm confused. I though if f was strict, then my program *always* > > printed 1? > > But not if you switch the (x <- ...) and (y <- ...) parts: > > main = do > r <- newIORef 0 > v <- unsafeInterleaveIO $ do > writeIORef r 1 > return 1 > y <- readIORef r > x <- case f v of > 0 -> return 0 > n -> return (n - 1) > print y > > Now the IORef is read before the case has a chance to trigger the writing.
But if the compiler is free to do this itself, what guarantee do I have that it won't? jcc _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
