Dear Marc, I ran this code: (c-declare "___SCMOBJ storage;") (define store! (c-lambda (scheme-object) void "storage = ___arg1;")) (define get (c-lambda () scheme-object "___return(storage);")) (define (gen-c) (let ((abc #f)) (##still-copy (lambda (v) (define o v) (set! abc v) o)))) (define gen-i (eval '(lambda () (let ((abc #f)) (##still-copy (lambda (v) (define o v) (set! abc v) o)))))) To test the hypothesis that a ____STILL closure can be successfully passed as a ___SCMOBJ from Scheme to C, stay in C for a while and then be moved back to Scheme anytime. A normal closure will not work, since it changes memory address. However, thanks to ##still-copy , I can get a closure that has a fixed memory address all up to collection, and such a closure's ___SCMOBJ can then be kept in C for as long as desired. Also, this ___STILL closure will still be collected, so there is no memory leak problem. The code above outputs:
(load "test") "/path/to/test.o1" (define x (gen-c)) (print x " ") (store! x) (##gc) (print (get) "\n") #<procedure #2 x> > > > #<procedure #2 x>
(define x (gen-i)) (print x " ") (store! x) (##gc) (print (get) "\n")
#<procedure #3 x> > > > #<procedure #3 x>
The only operational requirement for the transfer of a ___SCMOBJ from C to Scheme is that the object is still alive. In the example code above, that is guaranteed by that a strong reference to the closure is kept in |x| up to that (get) has made the transfer from C to Scheme, which then produces a new strong reference to the closure, which is sufficient in itself. Example to illustrate clean transfer of the strong reference: (define x (gen-c)) ; Generate closure (print x " ") (store! x) ; Store closure in C world, however still keep strong reference in |x| (##gc) ; All non-___STILL/___FIXED allocated objects will change ___SCMOBJ, thus invalidating them (define y (get)) ; We pick up the ___SCMOBJ from the C world and assign it to |y|. |x| above must be maintained as a strong reference up to the completion of this step. ; |x| is no longer needed for (get) to work, here, since we have a separate strong reference to the object now, in |y|. (set! x #f) (##gc) ; All fine. (print y "\n") The above example breaks, if the (##still-copy part is removed, in which case the closure will be ___MOVABLE, the ___SCMOBJ will be invalidated at GC, and therefore accessing the result of (get) produces SIGSEGV or similar unknown behavior, in my test it would print out #<unknown>. Marc please confirm that what I do above here is a best practice and is all correct. Phil