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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.
Let us consider the operation of the EPROM device in more detail. Consider the pining details below again Before we examine the interface means of the EPROM, it is worth
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