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Q. Explain about Deadlocks?
Deadlocks
for (int i = 0; i < n; i++) {
// first find a thread that can finish
for (int j = 0; j < n; j++) {
if (!finish[j]) {
boolean temp = true;
for (int k = 0; k < m; k++) {
if (need[j][k] > work[k])
temp = false;
}
if (temp) { // if this thread can finish
finish[j] = true;
for (int x = 0; x < m; x++)
work[x] += work[j][x];
At present there are two resources available. It is an system is in an unsafe state as process P1 could complete thereby freeing a total of four resources. But we can't guarantee that processes P0 and P2 can complete. Nevertheless it is possible that a process may release resources before requesting any further. For instance process P2 could release a resource thereby increasing the total number of resources to five. This permit process P0 to complete which would free a total of nine resources thereby allowing process P2 to complete as well.
What are a safe state and an unsafe state? Answer: A state is safe if the system can allocate resources to every process in some order and still avoid a deadlock. A system is
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Now let's implement bounded producer-consumer with semaphores, a Coke machine which is initially empty: semaphore sem_mutex = 1 // binary semaphore for mutual exclusion, 1==unlo
General Graph Directory A serious trouble with using an acyclic graph structure is making sure that there are no cycles. When links are further added to an existing tree struc
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