On January 17, 2011 19:24:12 Jan-Willem Maessen wrote:
Join allows you to merge the C1--C2--C3 dynamically determined from A--B--C into the continuation passing string to get
A--B--C--C1--C2--C3--D--E ... (C1--C2--C3 is determined by A--B--C)
This is actually relatively clear from join's type
join :: Monad m => m (m a) -> m a
It takes a string of computations (as indicated by the outer m) that generates a string of computations generating an a (as indicated by the inner m a). It changes this into just a string of computations generating an a.
How can it do this? It must set things up to first run the given string of computations in order to determine the string of computations that give a. Once it gets this, it must then run it in order to get a.
join (CPS f) = CPS $ \k -> unCPS (f id) k
That is, when invoked (i.e., when given a continuation k), invoke f with the continuation id. This will cause f to return the inner computation it computes. Then invoke this returned computation by passing it k.
You guys are both right. It shouldn't be any reflection on join or the description I gave for it though. I simply screwed up by lifting operations out of the continuation without thinking about preserving callCC. Let me restate the last little bit of the above. --- join (CPS f) = CPS $ \k -> f (g -> unCPS g k) That is, when invoked (i.e., when given a continuation k), invoke f with a continuation that takes the computation g generated by f and invokes it next by passing it k. --- My mistake in simplifying this expression was assuming f ends with returning unCPS k0 k1 (letting me lift the unCPS transformation and application of k1 out of the lambda giving id). callCC allows f to not end with unCPS k0 k1. The net effect of lifting these operations out was that it stops f from being able to discard them (i.e., escape past the join). Well subtle, I'm not sure it was too subtle. It is pretty clearly reflected in the types newtype Cont r a = CPS { unCPS :: (a -> r) -> r } join :: Cont (Cont r a) (Cont r a) -> Cont r a join (CPS f) = CPS $ \k -> unCPS (f id) k vs join :: Cont r (Cont r a) -> Cont r a join (CPS f) = CPS $ \k -> f (g -> unCPS g k) Cheers! -Tyson PS: I don't have anything against bind. I just find join is also nice to have for more functional/applicative style programming. It is also conceptually nice as it is quite clear about what additional complexity a monad gives you.