Inter Task Communication Questions and Answers

Answer: C

Correct answer: circular wait

Coffman’s four conditions must hold together: mutual exclusion, hold and wait, no preemption, and a circular chain of tasks each waiting for a resource the next holds. Breaking any one prevents deadlock, and imposing a global lock ordering is the usual way of breaking the fourth.

Answer: D

Neither will release, neither can proceed, and nothing in the system will change: the classic deadly embrace. Livelock differs in that the tasks keep running and keep failing, and starvation in that progress is possible but never granted.

Answer: A

A global order makes a circular wait impossible, and it costs nothing at run time - only discipline at design time. Timeouts and retries detect and recover from deadlock rather than preventing it, and can degenerate into livelock.

Answer: B

Correct answer: continue to execute but make no progress

Livelocked tasks are running, responding to one another and consuming processor time, but the system as a whole advances not at all - two tasks each backing off and retrying in step, for example. Deadlocked tasks are blocked and consume nothing, which is why livelock is harder to spot in a profile.

Answer: C

Correct answer: higher priority work is always ready to run

A strict priority scheduler runs the highest priority ready task, so a low priority task starves whenever something above it is always ready. It is a scheduling outcome rather than a locking fault, and the remedies are ageing the priority, reserving a share of processor time, or reconsidering the priority assignment.

Answer: D

Correct answer: giving each lock a longer timeout

A longer timeout makes a deadlock last longer before it is detected; it does nothing to prevent one. The other three each break a Coffman condition: lock ordering breaks circular wait, message passing removes the shared resource, and holding fewer locks reduces hold and wait.

Answer: A

Correct answer: a queue written from the handler with a zero timeout

A queue carries the data as well as the signal, and the from-interrupt variant with zero timeout never blocks - which is essential, since a handler that blocks cannot be resumed. A mutex must not be taken in a handler at all, and polling a global variable adds latency and race conditions.

Answer: B

Blocking in a handler is impossible, so the only options are to drop the item or to overwrite one, and either way the loss should be counted so it can be seen. A persistent overflow count is the signal that the queue or the consumer needs attention.

Answer: C

Correct answer: 25 samples

\[ \frac{50\ \mathrm{ms}}{2\ \mathrm{ms}} = 25\ \mathrm{samples}. \] That is the minimum, and a real design would add margin for jitter in the consumer’s scheduling. Sizing to the exact figure leaves nothing for the case where the task runs a little late.

Answer: D

Twenty-five slots of four bytes is a hundred bytes, plus the queue's own control structure of perhaps eighty more. On a part with two kilobytes of RAM that is a fifth of the budget for one sensor, which is why queue sizing is done deliberately rather than generously.