I am stuck while solving a counting semaphore problem in Operating system subject.

S is a semaphore initialized to 5.

count = 0 (shared variable) 

Assume that the increment operation in line #7 is not atomic.

Now,

1.  int counter = 0;
2.  Semaphore S = init(5);
3.  void parop(void)
4.  {
5.         wait(S);
6.         wait(S);
7.         counter  ;
8.         signal(S);
9.         signal(S);
10.  }

If five threads execute the function parop concurrently, which of the following program behavior(s) is/are possible?

A. The value of counter is 5 after all the threads successfully complete the execution of parop

B. The value of counter is 1 after all the threads successfully complete the execution of parop

C. The value of counter is 0 after all the threads successfully complete the execution of parop

D. There is a deadlock involving all the threads

what i have understand till now is answer is A and D,because what if all process are executed one by one say(T1->T2->T3->T4->T5) and final value saved will be 5(so A is one of correct options) Now, why D,because what if all process execute line 5 before 6 and will get blocked.

Now, please can any one help me to understand why B is another correct answer. ?

Thanks in advance,

Hope to here from you soon

Any help will be highly appreciated.

CodePudding user response：

Imagine thread 1 gets to line 7 before any other thread, and line 7 is implemented as three instructions:

7_1:  load counter, %r0
7_2:  add  $1, %r0
7_3:  store %r0, counter

For some reason (eg. interrupt, preempted), thread 1 stops at instruction 7_2; so it has loaded the value 0 into register %r0.

Next, thread's 2..5 all run through this sequence, leaving counter at say 4.

Thread 1 is rescheduled, increments %r0 to the value 1 and stores it into counter.