Re: Volatile happens before question

On 1/18/2012 4:34 PM, Patricia Shanahan wrote:


<http://vimeo.com/3757991>

If I understand correctly:

x & y are global integers initialized to 0. r1 and r2 are temporary
variables.

Thread 1:

int r1 = x;
y = r1;

Thread 2:

int r2 = y;
x = r2;

What happens is during the read of x, it is not found in cache. So a bus
cycle is started to read the value of x, but at the same time the cpu
has nothing to do while waiting, so it speculates as to the value of x
and continues processing. Let's say the cpu speculates that x is 42.

r1 = 42;

Now since r1 is 42, when y is written it also gets a speculative write
of the value 42;

y = 42;

This doesn't actually go on the memory bus, it's held in an output queue
of writes.

Next, Thread 2 comes along and also tries to read y. It also can't and
like thread 1 decides to speculate on the value of y while waiting for
main memory. It speculates y is 42.

r2 = 42;

Next x is written from the value of r2.

x = 42;

At this point, the internal cpu bus sees a write of 42 to the memory
location of x and thinks its speculation was correct, cancels the memory
read, and commits the value of y to memory.

That's what I understood from that talk on Vimeo. I'm starting to wonder
though exactly what sort of problem Bartosz Milewski is describing
there. I thought he was describing what could actually happen in the
absence of synchronization; possibly he is speculating on some kind of
hardware issue however.