🚀 go-pugleaf

In my earlier post, I neglected to include a very important cut in my
loop.  I am the first to admit cluelessness in Prolog programming, so
perhaps someone out there can give me some help optimizing use of the
heap w/ the following Turbo Prolog loop (in repeat ... fail form):
...........................................................................
run(Event_In):-                             % run takes a member of a special
     call_eval(Event_In, Event_Out),        % domain as an argument.
     Event_Out = done,!.                    % loop finished when symbol 'done'
                                            % is returned.
call_eval(Event, Event).
call_eval(Event_In, Event_Done):-
     exec_event(Event_In, Event_Out),       % SEE BELOW FOR exec_event***
     call_eval(Event_Out, Event_Done).

Notes on flow patterns:
    call_eval/2 is (i,o), exec_event/2 is (i,o), run/1 is (i)

*** exec_event takes an event and ultimately returns a new event from the
    old one -- this is the root of my problem; I want to reuse the same
    heap space since I won't need to ever backtrack.  Using retract and
    assert makes the program run unacceptably slow; I guess due to the
    complexity of the event data structure.  Could I use ref variables
    or something like that?  My compiler insists that call_eval is non-
    deterministic, since putting a cut in the first instance of call_eval
   would be contrary to the idea of the repeat..fail.  But, wouldn't this
   be the source of my heap leakage problem?

   If you'd like to run this program yourself, insert a storage check in
   call_eval, and then write a body for exec_event like:

   exec_event(Event_In, Event_Out):-
      Event_Out = Event_In,!.

Tim G.