Hi > The following is one of the laws. > > (x >>= f) >>= g == x >>= (\v -> f v >>= g)
Or stated another way: (x >>= f) >>= g == x >>= (f >>= g) Now it should be easier to see that this is simply associativity. It's easy enough to violate, if you want to - but I don't have any nice simple examples to hand. Thanks Neil _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
