Semaphores_Implemenation

Question

Consider this implementation of semaphores b/w these $2$ processes :
PROCESS-1 :

V(s)

CS

P(S)

PROCESS-2:

P(s)

CS

V(S)

Assume that semaphore "s" is intialized to $0$ at initial stage .
Comment on these statments :

• Process -$1$ can starve$.$
• Process=$2$ can starve$.$

Comments

But P1 is already in CS ? Since it can't come out of it, that's why it is starvation?

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Yes ! for subsequent computation it cannot access the resource.here CS i missed that point.

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There is no starvation in process 1 and process 2

Answer 1

both processes can starve.

initially S=0, process P1 can only start the execution.

Process P1 can execute again and again without giving a chance to process P2. So P2 can starve.

Process P1 starts the execution, P1 preempts, P2 starts its execution, now P2 can keep on running without giving a chance to P1, hence P1 can also starve.