Re: [BUG RT] dump-capture kernel not executed for panic in interrupt context

[Joerg Vehlow]

Hi Peter

On 9/7/2020 6:23 PM, peterz@xxxxxxxxxxxxx wrote:
> > According to the original comment in __crash_kexec, the mutex was used to
> > prevent a sys_kexec_load, while crash_kexec is executed. Your proposed patch
> > does not lock the mutex in crash_kexec.
> Sure, but any mutex taker will (spin) wait for panic_cpu==CPU_INVALID.
> And if the mutex is already held, we'll not run __crash_kexec() just
> like the trylock() would do today.
Yes you are right, it should work.
> > This does not cover the original use
> > case anymore. The only thing that is protected now are two panicing cores at
> > the same time.
> I'm not following. AFAICT it does exactly what the old code did.
> Although maybe I didn't replace all kexec_mutex users, I now see that
> thing isn't static.
Same thing here.
> > Actually, this implementation feels even more hacky to me....
> It's more minimal ;-) It's simpler in that it only provides the required
> semantics (as I understand them) and does not attempt to implement a
> more general trylock() like primitive that isn't needed.
Here I cannot agree with you. There is a second trylock in kernel_kexec, 
that cannot
be protected using the panic_cpu, but it actually could still use 
mutex_trylock and check
the panic_cpu. This should work I guess:

int kexec_trylock(void) {
    if (!mutex_trylock(&kexec_mutex)) {
        return 0;
    }
    smp_mb();
    if (panic_cpu != PANIC_CPU_INVALID) {
         mutex_unlock(&kexec_mutex);
         return 0;
    }
    return 1;
}

Or do I miss something now? All functions protected by mutex_lock cannot 
be executed, after
kexec_trylock resturned 1. kexec_crash will execute up to 
mutex_is_locked and then roll back.
The only thing that can go wrong now is: kexec_trylock executes up to 
smb_mb. At the same time
kexec_crash executes mutex_is_locked, which returns false now and then 
before panic_cpu is reset,
kexec_trylock executes the panic_cpu check, and returns. Now both 
functions did not get the lock and
nothing is executed.

Does that sound right to you? If you have no further objections I will 
post it here

Jörg