[Maxima-bugs] [ maxima-Bugs-752067 ] Can't untrace ?meval

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Bugs item #752067, was opened at 2003-06-10 19:59
Message generated for change (Comment added) made by crategus
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Category: Lisp Core
Group: None
Status: Open
Resolution: None
Priority: 2
Private: No
Submitted By: Stavros Macrakis (macrakis)
Assigned to: Nobody/Anonymous (nobody)
Summary: Can't untrace ?meval

Initial Comment:
If you trace(?meval), you can't untrace it!  This is 
presumably because you are running inside ?meval when 
untrace tries to untrace, but that's no excuse!

(The 22's below are just a trivial case of evaluation.)

(C1) 22;
(D1) 				      22
(C2) trace(?meval);
(D2) 				    [MEVAL]
(C3) 22;
1 Enter ?MEVAL [22]
1 Exit  ?MEVAL 22
(D3) 				      22
(C4) untrace(?meval);
1 Enter ?MEVAL [UNTRACE(MEVAL)]
1 Exit  ?MEVAL [MEVAL]
(D4) 				    [MEVAL]
(C5) 22;
1 Enter ?MEVAL [22]
1 Exit  ?MEVAL 22
(D5) 				      22
(C6) untrace();
1 Enter ?MEVAL [UNTRACE()]
1 Exit  ?MEVAL [MEVAL]
(D6) 				    [MEVAL]
(C7) 22;
1 Enter ?MEVAL [22]
1 Exit  ?MEVAL 22
(D7) 				      22


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>Comment By: Dieter Kaiser (crategus)
Date: 2009-11-06 20:40

Message:
The bug occurs not only for meval but for any traced function which is
called while $untrace is executed.

Again the bug. We do a Lisp trace for the function trace-handler:

(%i2) :lisp (trace trace-handler)
;; Tracing function TRACE-HANDLER.
(TRACE-HANDLER)

(%i2) trace(?meval);
(%o2) [?meval]

When we try to untrace meval the routine trace-handler is called for the
function meval itself with an argument '(($untrace) meval):

(%i3) untrace(?meval);
1. Trace: (TRACE-HANDLER 'MEVAL '((($UNTRACE) MEVAL)))
1 Enter ?meval [untrace(?meval)]
1 Exit  ?meval [?meval]
1. Trace: TRACE-HANDLER ==> ((MLIST SIMP) MEVAL)
(%o3) [?meval]

We trace in addition the function meval1. Now we can observe a call for
meval1 with the argument '(($untrace) meval1):

(%i4) trace(?meval1);
1. Trace: (TRACE-HANDLER 'MEVAL '((($TRACE) MEVAL1)))
1 Enter ?meval [trace(?meval1)]
1 Exit  ?meval [?meval1]
1. Trace: TRACE-HANDLER ==> ((MLIST SIMP) MEVAL1)
(%o4) [?meval1]

(%i5) untrace(?meval1);
1. Trace: (TRACE-HANDLER 'MEVAL '((($UNTRACE) MEVAL1)))
1 Enter ?meval [untrace(?meval1)]
2. Trace: (TRACE-HANDLER 'MEVAL1 '((($UNTRACE) MEVAL1)))
 1 Enter ?meval1 [untrace(?meval1)]
 1 Exit  ?meval1 [?meval1]
2. Trace: TRACE-HANDLER ==> ((MLIST) MEVAL1)
1 Exit  ?meval [?meval1]
1. Trace: TRACE-HANDLER ==> ((MLIST SIMP) MEVAL1)
(%o5) [?meval1]

This is the code to avoid the bug. We check if we try to untrace the
function fun:

(defun trace-handler (fun largs)
  (if (or return-to-trace-handle
          (and (not (atom (car largs)))
               (not (atom (caar largs)))
               (eq (caaar largs) '$untrace)
               (eq (cadar largs) fun)))
      ;; We were called by the trace-handler or by $untrace and the
function
      ;; fun is to be untraced.
      (trace-apply fun largs)
      ...

With this change it works as expected:

(%i7) untrace(?meval1);
1 Enter ?meval [untrace(?meval1)]
1 Exit  ?meval [?meval1]
(%o7) [?meval1]
(%i8) untrace(?meval);
(%o8) [?meval]
(%i9) trace;
(%o9) []

Dieter Kaiser

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