/// <summary>
/// Shows that it is possible for a shared variable that is modified without
/// synchronization to become corrupted. In this case, a counter is incremented
/// a certain number of times in parallel, but the resulting value is not
/// necessarily what we expect.
/// </summary>
private static void TheNeedForSynchronization()
{
Counter globalCounter = new Counter();
for (int i = 0; i < PARALLELISM; ++i)
{
ThreadPool.QueueUserWorkItem(IncrementCounterLotsOfTimes, globalCounter);
}
Thread.Sleep(TimeSpan.FromSeconds(2));
Console.WriteLine("Expected: {0}, actual: {1}", PARALLELISM * INCREMENTS, globalCounter.Current);
}
/// <summary>
/// Shows how a deadlock might ensue if resources are not locked in the same order.
/// In this example, thread #1 acquires locks in the order c1, c2 and thread #2
/// acquires locks in the order c2, c1. As a result, if sufficient time elapses,
/// the threads are deadlocked because they are waiting for each other. Detecting
/// and resolving deadlocks is outside the scope of this demo, but one example strategy
/// which prevents deadlocks such as this one is to perform lock ordering - always
/// acquire locks in the same order.
/// </summary>
private static void Deadlocks()
{
Counter c1 = new Counter();
Counter c2 = new Counter();
ThreadPool.QueueUserWorkItem(delegate
{
lock (c1)
{
Thread.Sleep(500);
for (int i = 0; i < 100000; ++i) c1.Next();
lock (c2)
{
for (int i = 0; i < 100000; ++i) c2.Next();
}
}
Console.WriteLine("Thread 1 done");
});
ThreadPool.QueueUserWorkItem(delegate
{
lock (c2)
{
Thread.Sleep(500);
for (int i = 0; i < 100000; ++i) c2.Next();
lock (c1)
{
for (int i = 0; i < 100000; ++i) c1.Next();
}
}
Console.WriteLine("Thread 2 done");
});
Console.Write("Wait a few seconds for the threads to be done. If nothing is printed, there's a deadlock");
Console.ReadLine();
}