Hi Elad,
On Wed, 11 Jan 2023 11:09:14 -0500, Elad Lahav wrote:
> Hello all,
>
> Here's an interesting (hopefully!) question I came across. Consider
> the following code:
>
> int a = 0;
> int b = 0;
>
> void
> writer()
> {
> a = 1;
> write_barrier();
> b = 2;
> }
>
> int
> reader1()
> {
> for (;;) {
> if (b == 2) {
> return a;
> }
>
> yield();
> }
> }
>
> int
> reader2()
> {
> while (b == 2) {
> yield();
> }
>
> return a;
> }
>
> The barrier issued by the writer ensures that if any reader observes b
> == 2 then it must also observe a == 1. However, that is only true if
> both the compiler and the processor agree that a is loaded after b.
>
> Consider the two reader functions. These are pretty much the same from
> C's point of view, but while the dependency between b and a is clear
> in the first case it is less so in the second. I can see a compiler
> loading a before the loop in reader2(), but less so in the case of
> reader1(). Perhaps that's just the way I interpret the code and a
> compiler would treat these as exactly the same.
>
> Let's assume then that you change the code to do something like return
> READ_ONCE(a) in both cases, as the book recommends. Does it change
> anything for the compiler? Does it change anything for the processor,
> which can reorder the reads? Anecdotally, it seems that a read barrier
> helps with reader2(), but I would not expect it to be needed for
> reader1().
>
> Thoughts?
I think Section 15.2.5 "Control Dependencies" can answer your question.
TLDR, control dependencies are good for load-to-store ordering.
You are talking about load-to-load ordering and control dependencies
don't suffice.
Thanks, Akira
>
> --Elad
