John Gateley

Solution to (call/cc call/cc)

The question: what does (call/cc call/cc) do in some program context, for example, something like:
(some-function arg1 (call/cc call/cc) arg3)

Approach 1 (a very inexact approach)

There's really only one continuation here, and there's not a lot you can do with that one continuation except return it, so a blind guess that is suprisingly correct is (call/cc (lambda (k) k)). That is, it just grabs the continuation and returns it.

Approach 2 (a more exact approach)

Use a calculus or semantics with rewrite rules. Here I use a meaning function E which takes an expression, an environment(ρ), a store(σ) and a continuation(κ); a denotational style approach, but not very rigorous. For clarity, I enumerate the two uses of call/cc: (call/cc1 call/cc2).
E[(call/cc1 call/cc2)]ρσκ
Evaluating call/cc1, we grab the continuation and pass it to the argument call/cc2
E[(call/cc2 κ)]ρσκ
Evaluating this second call/cc2 we grab the continuation and pass it to itself
E[(κ κ)]ρσκ
(κ κ)
Again, we discover that this is equivalent to (call/cc (lambda (k) k)):
E[(call/cc (lambda (k) k))]ρσκ
E[((lambda(k) k) κ)}ρσκ
(κ κ)