Re: [PATCH] let WARN_ON() output the condition

[Jiri Kosina]

On Wed, 6 Dec 2006, Ingo Molnar wrote:

> >                 DEBUG_LOCKS_WARN_ON(in_interrupt());    \
> >                 local_irq_save(flags);                  \
> >                 __raw_spin_lock(&(lock)->raw_lock);     \
> >                 DEBUG_LOCKS_WARN_ON(l->magic != l);     \
> > When one of these two WARN_ONs trigger, we get only
> > 	WARNING at kernel/mutex.c:132 __mutex_lock_common()
> no, that's not all we get - we should also get a stackdump. Are you not 
> getting a stackdump perhaps?

I am getting stackump, but I am perhaps just blind and don't see how to 
use it to distinguish the two WARN_ONs() conveniently, besides of 
disassembling the __mutex_lock_dommon and comparing it with offset in a 
stackdump. Well, not that it's not doable, but ...

> but i agree with you in theory that your proposed output is better, but 
> the side-effect issue is a killer i think. Could you try to rework it to 
> not evaluate the condition twice and to make it dependent on 
> CONFIG_DEBUG_BUGVERBOSE? You can avoid the evaluation side-effect issue 
> by doing something like:
> 	int __c = (c);							\
>                                                                         \
>         if (unlikely(__c)) {                                            \
>                 if (debug_locks_off())                                  \
>                         WARN_ON(__c);                                   \
>                 __ret = 1;                                              \
> 

Yep, making it dependent on CONFIG_DEBUG_BUGVERBOSE makes sense. Andrew, 
will you take such patch? (when I also fix the evaluating-twice issue).

Thanks,

-- 
Jiri Kosina
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