Creighton,
Yes, I would expect this example to run fine, not lock up.
Thanks again for diving into this non-termination problem.
- Conal
2008/11/17 Creighton Hogg
So in the interest of simplifying the recursive integral problem as much as possible, to make it a bit more tractable, I've come up with another recursive Behavior definition that I think should work if the current implementation of integral is expected to work. I'd just like to get an opinion on that to see if it's a test case I should try digging harder into. As I said, I _think_ that this should work iff the current definition of integral should work but if that's not true, please let me know. My reasoning is that this should be the equivalent of using integral with an event stream that has a constant difference of 1.
Thanks, Creighton
myIntegral :: Event a -> Behavior Double -> Behavior Double myIntegral e b = let e' = snapshot_ e b in sumB e'
x :: Event () -> Behavior Double x e = let b =myIntegral' e b in b
events :: Event () events = atTimes [0.1,0.2]
schedule a = threadDelay 100000 >> a
main = do c <- makeClock u <- mkUpdater (cGetTime c) (fmap print $ x events) forever $ schedule u
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