All~ I think I may know a way around this problem. You will have to bear with me for a moment as I am not entirely used to speaking in terms of continuations (I find a similar difficulty in speaking about Math, but at least I have a commonly accepted lexicon for that ;-).
The view from 10,000 feet. Full continuations do not operate on their context in place, but operate on a copy of it when invoked. The nitty gritty, consider (as provided by Jeff): a = 1 foo() print a b = 10 return b in the case where foo does not construct a full continuation an donly uses the return continuation, no extra copying is done and everything works. In the case where foo takes a full continuation and puts it in a global "evil", when foo upgrades its return continuation to a full one (which happens automatically if foo or one of its children creates a full continuation of its own), all of the continuations down the tree will be marked as full. When a full continuation is invoked it *copies* its contenxt into place (thus it can be invoked multiple times and it will always have its original context). This means that invoking full continuations will have a speed hit associated with it (although that is to be expected), creatings full continuations has a smaller speed hit of marking the chain (also reasonably expected), but invoking and creating return continuations would still be cheap. Hopefully that made sense to someone other than me, Matt -- "Computer Science is merely the post-Turing Decline of Formal Systems Theory." -???
