Re: Compiling C++ modules

Dmitry Torokhov
2006-04-25T18:04:10Z

On 4/25/06, Avi Kivity <avi@argo.co.il> wrote:
> Dmitry Torokhov wrote:
> >>>>>
> >>>> No, it's optimized out. gcc notices that &lock doesn't change and that
> >>>> 'l' never escapes the function.
> >>>>
> >>>>
> >>> "l" that propects critical section gets thrown away???
> >>>
> >> Calm down, the storage for 'l' is thrown away, but its effects remain.
> >>
> >
> > Would you mind explaining implemenation details a little bit?
> >
> (I don't know how familiar you are with C++ so I'm explaining it from
> the basics, apologies if I'm repeating things you know)
>
> Very often one needs to acquire a resource, do something with it, and
> then free the resource. Here, "resource" can mean a file descriptor, a
> reference into a reference counted object, or, in our case, a spinlock.
> And we want "free" to mean "free no matter what", e.g. on a normal path
> or an exception path.
>
> In C++, you code it as a guard object:
>
> struct spinlock_guard {
> spinlock_guard(spinlock_t *lock) { sl = lock; spin_lock(sl); }
> ~spinlock_guard() { spin_unlock(sl); }
>
> spinlock_t *sl;
> };
>

Oh, I remember now - lock is external wrt the code block it protects
so you just ensure destructor is called at the exit w/o allocating any
actual storage.
Thanks.

--
Dmitry