> Ok, so your example I translate to
>
> let fun f () = if ...
> then h ()
> else g ()
> and g () = if ...
> then h ()
> else f ()
> and h () = ...
> in K (f ())
> end
>
> where I filled in something to make the calls actually be tail calls.
>
> In this case, if you do NOT contify f or g, then I don't see how you can
> contify h. Would would it mean to do so since f and g each want to get to
> it.
I agree. You cannot contify h without contifying both f and g. The problem
with your original analysis is h will not be contified because its continuation
set is the two element set
{"return to f", "return to g"}
Fixing the analysis so that f's, g's, and h's continuation set are all the
singleton {K} while getting all the previous examples is what I'm working on.
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