Review: Monitor Semantics If P does X.wait() and later Q does X.signal(): –Hoare Semantics: Q blocks yielding the monitor immediately to P –Mesa Semantics: Q continues. P becomes runnable. P will eventually run, but only after Q exits –Brinch Hanson: Q can only signal as it exits, so P will run next. –Java synchrnoized methodsj: Only one reason to wait can exist. Notify() (the Java equivalent of signal() ) makes the blocked thread runnable as with Mesa semantics – it will run at a later time, after Q exits the monitor.

MONITOR prod-con { struct something buffer[n]; int in=0, out=0; int count=0; condition=full, empty; entry void add_item(data) { /* Mesa semantics require a while – why? Bounded wait? */ if/while (count==n) full.wait(); buffer[in] = data; in = (in + 1) % n; count++; empty.signal(); } /* continued */ Mesa vs. Hoare

Mesa vs. Hoare (cont.) /* continued */ entry remove_item (data_ptr) { /* Mesa semantics require a while – why? Bounded wait? */ if/while (count == n) empty.wait(); *data_ptr = buffer[out]; out = (out + 1) % n; count--; full.signal(); }

Dining philosophers P3P3 P1P1 P2P2 P0P0 P4P4 Rice Rules Philosophers alternate between thinking and eating 2 chopsticks are required to eat Philosophers never grab with both hands – they reach for one chopstick at a time. A philosopher is too polite to steal a chopstick from a colleague The philosophers cannot be allowed to starve More than one philosopher should be able to eat at a time (no token ring!)

Dining Philosophers: Approach 1 #define left(i) (i) #define right(i) ((i-1) % 5) MONITOR fork { int avail[5] = {2, 2, 2, 2, 2 }; /* forks available to each phil */ condition hungry[5]; entry pickup_fork (int phil) { if (avail[i] != 2) ready[i].wait(); avail[left(I)]--; avail[right(I)]--; } entry putdown_fork (int phil) { avail[left(i)]++: avail[right(i)]++; if (avail[left(i)] == 2) ready[left(i).signal(); if (avail[right(i)] == 2) ready[right(i)].signal(); }

Why Starvation? Each philosopher is waiting for a different condition, so we can’t ensure fairness. The solution is simple: all philosophers should wait for the same condition, such as their turn in a common queue.

Dining Philosophers: Approach 2 Semaphore chopstick[5] = { 1, 1, 1, 1, 1 }; while (1) { P(chopstick[i]); P(chopstick[(i+1) % 5]); >> V(chopstick[i]); V(chopstick[(i +1) % 5]); >> }

What is Deadlock? First, let’s define a resource: Resources are an abstraction of any reason to wait. Resources come in different types and we can have different numbers of each type. A process/thread can acquire a resource, use it, and then fee it. In this context, resources are unshareable – they are serially reusable. Now let’s formally define deadlock: The condition that arises when there exists a set of processes (or threads) such that each process holds a resource that another process in the set is waiting to acquire. The situation forces all processes in the set to wait forever.

Why Deadlock? Deadlock can occur if these conditions are satisfied: Mutual exclusion – at least one resource must be held by a process. Hold and wait – at least one process hold a resource while it is waiting for another resource. No preemption – one process can’t take another process’s resources in order to make progress (nor can the OS) Circular wait – there exists a circular chain of processes, each of which is waiting for a resource held by the next process in the chain.

Simple Defense: Serialization One simple defense against deadlock is to serialize the request of resources. 1.Enumerate all of the resources, giving each a number. 2.Require that all processes request resources in the order of this enumeration. That is to say that they are designed so that they never request a resource with a lower number than the highest numbered resource that they hold 3.Circular wait is now impossible, because the chain of waiting cannot wrap around from the greatest back to the beginning. It will eventually unfold.

Dining philosophers - Serialized P3P3 P1P1 P2P2 P0P0 P4P4 Rice Trace P 0 grabs Chopstick 0, blocking P 4 P 1 grabs Chopstick 1, blocking P 0 P 2 grabs Chopstick 2, blocking P 1 P 3 grabs Chopstick 3, blocking P 2 P 4 blocks trying to grab Chopstick 0 P 3 is free to grab Chopstick 4 P 3 can eat!

Dining Philosophers: Approach 3 Semaphore chopstick[5] = { 1, 1, 1, 1, 1 }; while (1) { if (i < ((i+1) % 5)) { P(chopstick[i]); P(chopstick[(i+1) % 5]); } else { P(chopstick[(i+1) % 5]); P (chopstick[i]); } >> V(chopstick[i]); V(chopstick[(i +1) % 5]); >> }