Question

Each of a set of n processes executes the following code using two semaphores a and b initialized to 1 and 0, respectively. Assume that count is a shared variable initialized to 0 and not used in CODE SECTION P.

CODE SECTION P
wait(a); count=count+1;
if (count==n) signal(b);
signal(a); wait(b); signal(b);
CODE SECTION Q

What does the code achieve?

Accepted Answer

The correct answer is (A) It ensures that no process executes CODE SECTION Q before every process has finished CODE SECTION P.. This is the classic semaphore barrier pattern. Semaphore a mutually excludes access to 'count' while incrementing it (mutual exclusion during the counting step, not throughout all of Section P). The LAST process to arrive (whose increment makes count==n) signals b once. Every process then does wait(b) — the first n−1 processes block here until that final signal arrives, and each process that passes wait(b) immediately does signal(b) again, 'passing along' exactly one permit so the next waiting process can also proceed (a classic cascading-release barrier). The net effect: NO process can enter CODE SECTION Q until every one of the n processes has incremented count, i.e. until all have reached (finished) CODE SECTION P — a barrier synchronization.

Explanation