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let S and Q to be initialized to 1 

process p0:          process p1:

wait(S);             wait(Q);
wait(Q);             wait(S); 

..                    ..
..                    ..
Signal(S);           Signal(Q);
Signal(Q);           Signal(S);

I know there is a deadlock , where both the process will be going into the sleep , so what about starvation and bounded waiting?

My approach:

as only p0 can be executed any number of times i think there will be starvation and no bounded waiting.But answer is given that there is No Starvation.

Pavan Kumar Munnam
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1 Answers1

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The answer is wrong.

Deadlock implies starvation, and the given code suffers from deadlock, here's how.

Say p0 executes wait(S); and then p1 executes wait(Q);. Now both the semaphores S and Q have value = 0.Now p0 will keep on waiting due to statement wait(Q); and p1 will keep on waiting due to statement wait(S);.

Freedom from starvation means that:

if a thread tries to acquire a lock , it will eventually succeed.

This does not happen here, so the answer is wrong.

Sumeet
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