Re: [Haskell-cafe] Re: questions about Arrows

On Wed, 2010-09-01 at 11:49 -0700, Ben wrote:

Thanks for the prompt reply. Some questions / comments below :

On Wed, Sep 1, 2010 at 12:33 AM, Maciej Piechotka wrote:

...

rSum2 :: ArrowCircuit a => a Int Int rSum2 = proc x -> do rec out <- delay 0 -< out + x returnA -< out + x

Wow, that was simple. I guess I never thought to do this because it evaluates (out + x) twice, but one can always write

rSum3 :: ArrowCircuit a => a Int Int rSum3 = proc x -> do rec let next = out + x out <- delay 0 -< next returnA -< next

I have a follow-up question which I'll ask in a new thread.

Possibly it should be written as rSum4 :: ArrowCircuit a => a Int Int rSum4 = proc x -> do rec let !next = out + x out <- delay 0 -< next returnA -< next

...

...

3) One can define fix in terms of trace and trace in terms of fix.

trace f x = fst $ fix (\(m, z) -> f (x, z)) fix f = trace (\(x, y) -> (f y, f y)) undefined

Does this mean we can translate arbitrary recursive functions into ArrowLoop equivalents?

Yes. In fact fix is used on functional languages that do not support recursion to have recursion (or so I heard)

Ups. Sorry - my dyslexia came to me and I read recursive instead of ArrowLoop. My fault IMHO it is not possible.

In which case my question is, why is the primitive for Arrows based on trace instead of fix?

How would you define loop in terms of fixA :: ArrowLoop a => a b b -> a c b fixA f = loop (second f >>> arr (\(_, x) -> (x, x))) The only way that comes to my mind is: loopA :: (ArrowLoop a, ArrowApply a) => a (b, d) (c, d) -> a b c loopA f = proc (x) -> do arr fst <<< fixA (proc (m, z) -> do f -< (x, z)) -<< x Which requires ArrowApply. So as long as arrow is ArrowLoop and ArrowApply it is possible.

Best regards, Ben

Regards