Re: Declaring unrecoverable_exception() as __noreturn ?

[Christophe Leroy <christophe.leroy@csgroup.eu>]

Le 11/02/2021 à 05:41, Michael Ellerman a écrit :
> Nicholas Piggin <npiggin@gmail.com> writes:
>> Excerpts from Christophe Leroy's message of February 11, 2021 2:44 am:
>>> As far as I can see, almost all callers of unrecoverable_exception() expect it to never return.
>>>
>>> Can we mark it __noreturn ?
>>
>> I don't see why not, do_exit is noreturn. We could make die() noreturn
>> as well.
> 
> I'm always nervous about that, because we can return if a debugger is
> involved:
> 
> DEFINE_INTERRUPT_HANDLER(unrecoverable_exception)

Hum ... Is that correct to define it as an interrupt handler ?

Also, I see it declared a second time in interrupt.c, this time not as an interrupt handler. Is that 
wanted ?

> {
> 	pr_emerg("Unrecoverable exception %lx at %lx (msr=%lx)\n",
> 		 regs->trap, regs->nip, regs->msr);
> 	die("Unrecoverable exception", regs, SIGABRT);
> }
> 
> void die(const char *str, struct pt_regs *regs, long err)
> {
> 	unsigned long flags;
> 
> 	/*
> 	 * system_reset_excption handles debugger, crash dump, panic, for 0x100
> 	 */
> 	if (TRAP(regs) != 0x100) {
> 		if (debugger(regs))
> 			return;
> 
> 
> We obviously don't want to optimise for that case, but it worries me
> slightly if we're marking things noreturn when they can actually return.
> 

I don't think I want to declare die() as __noreturn, need to look at it more in details first.

Christophe