Peter Verswyvelen wrote:
Paul L wrote: A minor detail in your paper: on page 7, you represent *(d) sf1 &&& sf2 *as a big box taking one input and producing two outputs. The input is internally split using a Y. This does not seem consistent with the other boxes (e.g. *first *or *loop *internally) that show two arrows for an incoming/outgoing pair, so I would say the outer box of &&& would also have two inputs and two outputs.
But look at the type of &&&: (&&&) :: Arrow a => a b c -> a b c' -> a b (c, c') or, perhaps more readable, (&&&) :: Arrow (~>) => (b ~> c) -> (b ~> c') -> (b ~> (c, c')) As you can see, the resulting arrow of type (b ~> (c, c')) really has only one input and produces a pair, i.e. two outputs. Internally it must duplicate the b input somehow and apply it to both input arrows, exactly as the box shows. Bertram