In the following incomplete code for readers-writers problem, two binary semaphores mutex and wrt are used to obtain synchronization. wait (wrt) writing is performed signal (wrt) wait (mutex) readcount = readcount + 1 if readcount = 1 then S1 S2 reading is performed S3 readcount = readcount − 1 if readcount = 0 then S4 signal (mutex) The value of S1, S2, S3, S4, (in that order) are _____________
Wait (wrt), Signal (mutex), Wait (mutex), Signal (wrt)
In the first readers–writers solution, wrt gives writers exclusive access to the shared data and mutex protects the shared counter readcount. Writers simply do wait(wrt), write, signal(wrt). Readers coordinate so that the first reader locks writers out and the last reader lets them back in.
Step 1 — S1: after incrementing readcount under mutex, the first reader (readcount = 1) must block writers, so S1 is wait(wrt).
Step 2 — S2: before reading, the reader releases mutex so other readers can enter concurrently, so S2 is signal(mutex).
Step 3 — S3: after reading, the reader must lock the counter again before decrementing it, so S3 is wait(mutex).
Step 4 — S4: when the last reader leaves (readcount = 0) writers may proceed, so S4 is signal(wrt); the final signal(mutex) then releases the counter.
Any sequence that begins with a signal on mutex would release a semaphore that the reader does not yet hold at that point, and a sequence ending in wait(wrt) would leave writers blocked forever, so the other orders break mutual exclusion or cause starvation of writers.
Wait (wrt), Signal (mutex), Wait (mutex), Signal (wrt) option is correct.
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