Re: Signal delivery order

[Oleg Nesterov <oleg@redhat.com>]

On 03/15, Gábor Melis wrote:
>
> On Domingo 15 Marzo 2009, Oleg Nesterov wrote:
> >
> > If test_signal (SIGUSR1) is blocked, this means it is already
> > delivered, and the handler will be invoked when we return from
> > sigsegv_handler(), please see below.
>
> SIGUSR1 is delivered, its sigmask is added to the current mask but the
> handler is not yet invoked and in this instant synchronous sigsegv is
> delivered, its handler invoked?

Can't understand the question. Could you reiterate?

> > When sigprocmask(SIG_UNBLOCK) returns, both signals are delivered.
> > The kernel deques 1 first, then 2. This means that the handler for
> > "2" will be called first.
>
> My mental model that matches what I quickly glean from the sources (from 
> kernel/signal.c, arch/x86/kernel/signal_32.c) goes like this:
>
> - signal 1 and signal 2 are generated and made pending
> - they are unblocked by sigprocmask
> - signal 1 is delivered: signals in its mask (only itself here) are
> blocked

yes.

the kernel changes ip (instruction pointer) to sig_1.

> its handler is invoked

no.

We never return to user-space with a pending signal. We dequeue signal 2
too, and change ip to sig_2.

Now, since there are no more pending signals, we return to the user space,
and start sig_2().

> I can't find how 'handler for "2" will be called first'.

see above,

> Furthermore, if
> it's indeed sig_2 that's invoked first then sig_1's sigmask is added to
> the current mask upon dequeueing???

sig_1's sigmask was added to ->blocked when we dequeued signal 1.

Oleg.