Last night I found "A Better API for First-Class Continuations", which
explains how a debugger can take control and give it back to the original
computation. The paper also describes the tree created by a series of
continuations. Is there a way to see the whole tree? I'd like to under-
stand how the backtrace command creates the stack out of the tree.
You need a first error or step to show a backtrace. As the transcript
below shows, a second error will change the backtrace (I had hoped it would
add frames to the printout). Curiously the change is only temporary --
"undoing" the second error with ,d also undoes the change. As the second
test shows, tail recursion is not involved. I assume the "change" is really
an effect of the ,b printing code.
Thanks!
-- Derek
Gambit v4.2.4
> (define (a) ; a (step) in tail position doesn't cause a stop -- doubling
(step) (step)) ; creates a non-tail (step) which is what the test uses
> (a)
*** STOPPED IN a, (console)(a)2.9
1> ,b
0 a (console)@2:9 step
1 (interaction) (console)@3:1 (a)
1> x
*** ERROR IN (console)(a)5.1 -- Unbound variable: x
2> ,b
0 (interaction) (console)@5:1 x
1 (interaction) (console)@3:1 (a)
2> ,d
1> ,b
0 a (console)@2:9 step
1 (interaction) (console)@3:1 (a)
1> ,q
Gambit v4.2.4
> (generate-proper-tail-calls #f)
> (define (a)
(step) (step))
> (a)
*** STOPPED IN a, (console)(a)3.9
1> ,b
0 a (console)@3:9 step
1 (interaction) (console)@4:1 (a)
1> x
*** ERROR IN (console)(a)6.1 -- Unbound variable: x
2> ,b
0 (interaction) (console)@6:1 x
1 (interaction) (console)@4:1 (a)
2> ,d
1> ,b
0 a (console)@3:9 step
1 (interaction) (console)@4:1 (a)
1> ,q