(define (f x) (let lp ((i 0)) (if (< i 10) (begin (pk 'value-from-parent (pause x i)) (lp (+ i 1)))))) (define (test-1) (let ((x (make-pause-stack)) (ret 0)) (let lp ((i 0)) (let-values (((k x) (resume x (- i)))) (cond ((= k pause) (pk 'value-from-child x) (lp (+ i 1))) ((= k parent) (pk 'parent)) ((= k leave) (pk 'leave)) ((= k child) (pk 'child) (f x) (leave x) (set! ret i))))) (pk 'finish x) ret)) On Sun, Feb 13, 2022 at 11:27 AM Mikael Djurfeldt wrote: > Hi, > > I'm trying to understand this. > > The example of a generator which you give below counts upwards, but I > don't see how the value of n is passed out of the generator. > > Could you give another example of a generator which does pass out the > values, along with a usage case which prints out the values returned by the > generator? > > Best regards, > Mikael > > Den tors 10 feb. 2022 17:52Stefan Israelsson Tampe < > stefan.itampe@gmail.com> skrev: > >> Consider a memory barrier idiom constructed from >> 0, (mk-stack) >> 1. (enter x) >> 2. (pause x) >> 3. (leave x) >> >> The idea is that we create a separate stack object and when entering it, >> we will swap the current stack with the one in the argument saving the >> current stack in x and be in the 'child' state and move to a paused >> position in case of a pause, when pausing stack x, we will return to where >> after where entered saving the current position in stack and ip, and be in >> state 'pause' and when we leave we will be in the state 'leave and move >> to the old stack, using the current >> ip. At first encounter the function stack frame is copied over hence >> there will be a fork limited to the function only. >> >> This means that we essentially can define a generator as >> (define (g x) >> (let lp ((n 0)) >> (if (< n 10) >> (begin >> (pause x) >> (lp (+ n 1)))))) >> >> And use it as >> (define (test) >> (let ((x (mk-stack))) >> (let lp () >> (case (enter x) >> ((pause) >> (pk 'pause) >> (lp)) >> ((child) >> (g x) >> (leave x)))))))) >> >> A paused or leaved stack cannot be paused, an entered stack cannot be >> entered and one cannot leave a paused stack, but enter a leaved stack. >> >> Anyhow this idea is modeled like a fork command instead of functional and >> have the benefit over delimited continuations that one does not need to >> copy the whole stack and potentially speed up generator like constructs. >> But not only this, writing efficient prolog code is possible as well. We >> could simplify a lot of the generation of prolog code, speed it up and also >> improve compiler speed of prolog code significantly. >> >> How would we approach the prolog code. The simplest system is to use >> return the >> alternate pause stack when succeeding things becomes very simple, >> >> x = stack to pause to in case of failure >> cc = the continuation >> >> ( (x cc) goal1 goal2) >> :: (cc (goal1 (goal2 x)) >> >> ( (x cc) goal1 goal2) >> :: (let ((xx (mkstack))) >> (case (enter xx) >> ((child) >> (cc (goal2 xx))) >> >> ((pause) >> (cc (goal2 x))))) >> >> Very elegant, and we also can use some heuristics to store already made >> stacks when >> leaving a stack and reuse at the next enter which is a common theme in >> prolog, >> >> Anyhow we have an issue, consider the case where everythings >> succeds forever. Then we will blow the stack . There is no concept of tail >> calls here. So what you can do is the following for an , >> >> (let ((xx (mk-stack))) >> (case (enter xx) >> ((child) >> (goal1 x (lambda (xxx) (pause xx xxx))) >> >> ((pause xxx) >> (goal2 xxx cc)))) >> >> This enable cuts so that a cutted and (and!) in kanren lingo will use >> (goal2 x cc) >> >> And we have tail calls! >> >> >> I have a non jitted version guile working as a proof of concept. >> >> The drawback with this is if a function uses a lot of stack, it will be a >> memory hog. >> >> WDYT? >> >> >> >> >> >> >> >> >> >> >> >> . >> >> >> >