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ddb_onpanic [Re: CVS commit: src/sys/kern]
[ moved to tech-kern ]
On Sat, Jan 28, 2006 at 12:48:30PM +0000, Darren Reed wrote:
> > If the system is in such a state that ddb can't print its prompt,
> > it's not clear if db_stack_trace_print() will work better. Also, on some
> > platforms, ddb() may need some more initialisations before
> > db_stack_trace_print() will work (it's just theory, I've no example of
> > platforms where this would be the case).
> > Also, with your change there's no way to get the old ddb.onpanic=0
> > behavior: if ddb.onpanic is not 1, db_stack_trace_print will always be
> > called.
>
> Huh ?
>
> ddb.onpanic=0 functionality has not changed. For non-zero values
> that are neither 1 or 2, it has changed, but they're undefined.
>
> It previously did "if (db_onpanic)" and I made it "if (db_onpanic == 1)"
>
> The other "changes" were style...
>
> Or am I missing something ?
In panic() I see:
if (db_onpanic == 1)
Debugger();
else {
static int intrace = 0;
if (intrace == 0) {
intrace = 1;
printf("Begin traceback...\n");
db_stack_trace_print(
(db_expr_t)(intptr_t)__builtin_frame_address(0),
TRUE, 65535, "", printf);
printf("End traceback...\n");
intrace = 0;
} else
printf("Faulted in mid-traceback; aborting...");
if (db_onpanic == 2)
Debugger();
}
If db_onpanic is not 1, db_stack_trace_print() will be called, even for
db_onpanic == 0. There's no way to avoid this.
IMHO if we keep this the else should be changed to
else if (db_onpanic != 0)
I will reply to the second part of your mail separately (need to do a few
investigations first).
--
Manuel Bouyer <bouyer%antioche.eu.org@localhost>
NetBSD: 26 ans d'experience feront toujours la difference
--
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