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Re: (call/cc call/cc)
No, (callcc callcc) is not the identity continuation but the current
continuation. If E is an evaluation context, then E[(callcc callcc)]
is more or less E -
(lambda (x) (abort E[x])).
The abort is important.
Sorry I looked for the dvi or ps of my LFP 88 paper on such things and
couldn't find it. You will need to lok in your library:
M, Felleisen. lambda-v-CS: An extended lambda-calculus for
Scheme. In Proc. 1988 Conference on Lisp and Functional
Programming, 1988, 72--84.