Brian Denheyer <bri...@aracnet.com> writes: >>>>> doEvent f usDelay = forkIO $ >>>>> threadDelay usDelay >>>>> doEvent f usDelay >>>>> f
There's a missing 'do' here, right? >> Infinite loop? yes, that is what you wanted. Memory gobbling? Why >> would you think that? > Why would I think that ? > doEvent f usDelay = do forkIO $ threadDelay usDelay > doEvent f usDelay > f Are you sure this isn't interpreted as: doEvent f usDelay = do (forkIO $ threadDelay usDelay) doEvent f usDelay f I.e. just forking off processes that only does a delay, never even getting to 'f'? I interpreted Thomas's code to mean: doEvent f usDelay = forkIO $ (do threadDelay usDelay doEvent f usDelay f) Which appears to work nicely here. -k -- If I haven't seen further, it is by standing in the footprints of giants _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe