The solution is relatively simple:
- Philosopher attempts to get fork on left
SUCCEED -> Continue to step 2
FAIL -> wait (for a while)- Philosopher attempts to get fork on right
SUCCEED -> Continue to step 3
FAIL -> Release left fork and wait (for a while)- Eat and release both forks. Then wait (for a while)
The point to emphasize here is that anytime a philosopher fails to get both forks, he needs to drop any forks he is holding and wait a bit or deadlock will eventually occur.
Perhaps more importantly, what kind of moron uses two forks to eat?
— EDIT —
Here is a quick example for the Fork
class Fork {
public static final char FORK = '|';
public static final char NO_FORK = ' ';
private int id;
private Lock lock = new ReentrantLock();
public Fork(final int id) {
this.id = id;
}
public boolean isHeld() {
return lock.isLocked();
}
// returns true if successfully grabbed!
public synchronized boolean tryToGrab() {
return lock.tryLock();
}
public void letGo() {
lock.unlock();
}
}
Your idea of using a Semaphore object would work just as well. Good luck!
13
solved Deadlock occuring in the Dining Philosophers code