jmg-duarte / cp-tests Goto Github PK

Não compreendo como poderá ser possível este x = 4.

Bom trabalho e obrigado pela partilha destes docs para o teste.

If only two threads deadlock or livelock, shouldn't a third, doing a completely different computation, be able to progress?

If all threads do a simultaneous compare&swap, it will fail for all, I think. That's why you need to add backoffs.

Also, what is busy-waiting?

Thanks again for taking the time to help everyone out.

This solution does not have into account "y=3" is atomic.

So it needs to be completely remade.

T1	T2	T3		T1	T2	T3		T1	T2	T3		T1	T2	T3
read x=0				read x=0				read x=0				read x=0
	read y=0				read y=0					write y=3				write y=3
	write y=1				write y=1				read y=3				read y=3
		write y=3				write y=3			write y=4				write y=4
		read y=3		write x=1						read y=4				read y=4
		write x=4				read y=3				write x=5		write x=1
write x=1						write x=4		write x=1						write x=2

T1	T2	T3		T1	T2	T3		T1	T2	T3		T1	T2	T3
	read y=0					write y=3
	write y=1				read y=3
		write y=3			write y=4
		read y=3				read y=4
		write x=4				write x=5
read x=4				read x=5
write x=5				write x=6

X = 1,2,4,5,6
Y = 3,4
(X,Y) = (1,3), (1,4), (2,4), (4,3), (6,4), (5,3)

shouldn't these be marked? E(4,0) to E(4,3);

Don't we need to prove there is a common correlation for a high-level data race to exist? How do we go about doing this?

Acho que pode dar um deadlock quando fazes:
transferTo(acc[2], acc[1]);
acc[0].credit(interest(0.01);

No transferTo, vais tentar ter o lock da acc[1] que já tá bloqueada, e no interest vais tentar ter o lock acc[2] que já está bloqueado.

Teria de ser a ordem:
interest lock acc[1] -> transfer2 lock acc[2] -> Deadlock?

cumps

Apesar de um pouco semelhantes, julgo que esta será a E.

It's B. RAR's do not.

Shouldn't it be C?

From the slides:

Aqui só consegui obter: 3, 5, 6. Não consigo obter o 4.