On Mon, Sep 18, 2017 at 07:51:29AM +0900, Akira Yokosawa wrote:
> On 2017/09/17 14:55:08 -0700, Paul E. McKenney wrote:
> > On Sun, Sep 17, 2017 at 08:04:21PM +0900, Akira Yokosawa wrote:
> >> On 2017/09/16 18:07:30 -0700, Paul E. McKenney wrote:
> >>> On Sat, Sep 16, 2017 at 08:01:45PM +0900, Akira Yokosawa wrote:
> >>>> Hi Paul,
> >>>>
> >>>> I'm a bit disturbed by the description in Section 14.3.1 "Memory-Reference
> >>>> Restrictions" quoted below:
> >>>>
> >>>>> Oddly enough, the compiler is within its rights to use a variable
> >>>>> as temporary storage just before a store to that variable, thus
> >>>>> inventing stores to that variable.
> >>>>> Fortunately, most compilers avoid this sort of thing, at least outside
> >>>>> of stack variables.
> >>>>> Nevertheless, using WRITE_ONCE() (or declaring the variable
> >>>>> volatile) should prevent this sort of thing.
> >>>>> But take care: If you have a translation unit that uses that variable,
> >>>>> and never makes a volatile access to it, the compiler has no way of
> >>>>> knowing that it needs to be careful.
> >>>>
> >>>> I'm wondering if using WRITE_ONCE() in a translation unit is really
> >>>> enough to prevent invention of stores.
> >>>>
> >>>> Accessing via a volatile-cast pointer guarantees the access is not
> >>>> optimized out (and hopefully the referenced value is respected).
> >>>>
> >>>> But I suspect that it has any effect in preventing invention of extra
> >>>> loads/stores.
> >>>>
> >>>> Isn't declaring the variable volatile necessary for the guarantee?
> >>>>
> >>>> In practice, as is described in the above quote: "Fortunately, most
> >>>> compilers avoid this sort of thing, at least outside of stack variables",
> >>>> we can assume non-volatile shared variables are not spilled out to
> >>>> the variables themselves as far as GCC/LLVM is concerned.
> >>>> But this is compiler dependent, I suppose.
> >>>
> >>> I suspect that it will turn out to be impossible for the compiler to
> >>> actually invent these stores in the general case. For example, it might
> >>> be that there is some lock held or other synchronization mechanism unknown
> >>> to the compiler that prevents this behavior. But I haven't fully worked
> >>> this out yet.
> >>
> >> You mean the invented stores wouldn't be visible from other threads anyway?
> >> In a meaningful parallel code, that can be the case.
> >
> > I mean that it is very hard to prove that inventing a store isn't introducing
> > a data race, which would be a violation of the standard. The one case I know
> > of where the compiler can be sure that it is within its rights to invent the
> > store is before a normal store to a variable.
> >
> > Otherwise, it might be (for example) that one must hold a lock to legally
> > update a given variable, and that lock might or might not be held at a given
> > point in the code. But if the compiler sees a plain store, the compiler
> > knows that it is OK to update at that point. So the compiler can invent
> > a store prior to the existing store, as long as there is no memory barrier,
> > compiler barrier, lock acquisition/release, atomic operation, etc., between
> > the original store and the compiler's invented store.
>
> I think I understand.
>
> >
> >>> But I do know that if you just do plain stores, the compiler is fully
> >>> within its rights to invent stores preceding any given plain store.
> >>
> >> So, the rules to use WRITE_ONCE() is something like the following?
> >>
> >> ---
> >> 1) Declare the variable without volatile.
> >
> > Agreed.
> >
> >> 2) READ_ONCE() and plain loads can be mixed. A plain load will see
> >> a value at least newer than or equal to the one obtained at the
> >> program-order most recent READ_ONCE().
> >
> > I am not entirely sure of this one. But if there is a barrier() or
> > stronger between the READ_ONCE() and the plain load, then yes.
>
> Ah, the compiler can reorder plain loads before READ_ONCE()...
>
> I did a litmus test of a plain load after READ_ONCE(), but
> such a reordering is not covered by herd7's litmus test, is it?
>
> >
> >> 3) WRITE_ONCE() should not be mixed with plain stores when invention
> >> of stores is to be avoided.
> >
> > Agreed.
> >
> >> Invention of stores is the opposite of fusing stores.
> >> Suppose you don't want to update progress in the while loop:
> >>
> >> while (!am_done()) {
> >> do_something(p);
> >> tmp++;
> >> }
> >> progress = tmp;
> >>
> >> The compiler might transform this to
> >>
> >> while (!am_done()) {
> >> do_something(p);
> >> progress++;
> >> }
> >
> > But only as long as the compiler knows that do_something() doesn't
> > contain any ordering directives.
>
> Yes. I borrowed the fusing example in the text and it should have
> the same assumption.
>
> >
> >> if it wants to avoid allocation of a register/stack to tmp for whatever
> >> reason. WRITE_ONCE() prevents the unintended accesses of progress:
> >>
> >> while (!am_done()) {
> >> do_something(p);
> >> tmp++;
> >> }
> >> WRITE_ONCE(progress, tmp);
> >
> > Agreed, this would prevent the update to "progress" from being pulled
> > into the loop.
> >
> >> ---
> >> Adding this example in the text might be too verbose.
> >> Would a Quick Quiz be reasonable?
> >
> > Might be good in the section on protecting memory references, and putting
> > it into a quick quiz or two makes a lot of sense.
>
> It's up to you where to put it.
>
> And I now realize using READ_ONCE() and WRITE_ONCE() is quite tricky.
> Missing one might not cause a problem today, but a smarter compiler
> can expose the bug in the future...
>
> This is scary.
Section 15.3.1 is supposed to cover READ_ONCE() and WRITE_ONCE().
There is one final paragraph added just now, but if you get a chance,
please let me know what you think.
And if you are scared, you might actually have a good understanding
of the true situation. ;-)
Thanx, Paul
> Thanks, Akira
>
> >
> > Thanx, Paul
> >
> >> Thanks, Akira
> >>
> >>>
> >>> Thanx, Paul
> >>>
> >>>
> >>
> >
> >
>
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