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From: Elad Lahav <e2lahav@gmail.com>
To: perfbook@vger.kernel.org
Subject: Ordering and conditionals
Date: Wed, 11 Jan 2023 11:09:14 -0500	[thread overview]
Message-ID: <CAJbg=FX8rBFRqO-3bhE863OepcdShoyGg0tVeedVR2=tfjwcDQ@mail.gmail.com> (raw)

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?

--Elad

next             reply	other threads:[~2023-01-11 16:09 UTC|newest]

Thread overview: 2+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2023-01-11 16:09 Elad Lahav [this message]
2023-01-11 22:20 ` Ordering and conditionals Akira Yokosawa

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