The program isn't clear to me. I guess you want something like this:
int x;
volatile int v;
CPU1:
write(x, 1) (1)
write(v, 1) (2)
CPU2
read(v) (3)
read(x) (4)
Then there exists an execution where (4) sees value written by (1) it is in
a data race with. E.g. (1)->(3)->(4).
So since the program has at least 1 execution with a data race. So your
observation that the program is not data race free, is correct.
To resolve the data race, you want to do something like this:
int x;
volatile int v;
CPU1:
write(x, 1) (1)
write(v, 1) (2)
CPU2
if(read(v)==1){ (3)
print(read(x)); (4)
}
When (3) has observed the value written by (2) there is a happens-before
edge between (2) and (3). And due to the program order between (1)/(2) and
(3)/(4) and the transitive nature of happens-before, there is a
happens-before edge between (1) and (4). So the value'1' should be printed.
This program is data race free.
On Sun, Oct 23, 2022 at 7:59 PM r r <grosi...@gmail.com> wrote:
> int x;
> volatile int v;
>
> write(x, 1) read(v)
> write(v, 1) read(x)
>
> Is that program data race free? On my eye it is not not because there is
> an execution that write(x, 1) and read(x) are not in happens-before
> relationship.
>
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