JMM - is this program data race free?

[r r]

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.

[Peter Veentjer]

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.

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[r r]

Thanks a lot

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