Hi,
I assume you are worried that (1) can be moved by the compiler below
"executor.execute()" or on processors with weak memory models changes from
(1) might not be fully visible to other threads when (2) is executed. I
believe it is not going to happen, as (1) should happen before
"executor.execute()" (because they are on the same thread), which in turn
should happen before retrieving a task for execution (because they are
synchronised via task queue), which again happens before (2) (because on
the same thread with the retrieving operation).
Best regards,
Cezary
On Tue, Sep 25, 2018 at 5:52 PM John Hening wrote:
> public class Test {
> ArrayList xs;
> ArrayList doers;
> Executor executor = Executors.newSingleThreadExecutor();
>
> static class Doer {
> public void does(X x){
>x.f(); //
> (2)
> }
> }
>
> void test() {
> for(X x : xs){
> x.f(); //
> (1)
>
> for(Doer d : doers) {
> executor.execute(() -> d.does(x));
> }
> }
> }
> }
>
>
>
>
> For my eye, if X.f is not synchronized it is incorrect because of two
> facts (and only that two facts):
>
> 1. Obviously, there is data race between (1) and (2). There are no more
> data races here. (doers contains different objects)
> 2. There is no guarantee that (1) will be executed before (2). Yes?
>
> If X.f would be synchronized that code will be correct because:
> 1. There is no data race.
> 2. There is guarantee that (1) will be executed before (2) because (1) is
> a synchronization action and Executor.execute is also a synchronization
> access (not specifically execute itself)
>
> Yes?
>
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Cezary Biernacki
Director of Software Development
crosswordcybersecurity.com @crosswordcyber
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