Следующий код почти наверняка создаст взаимоблокировку и продемонстрирует классический сценарий взаимоблокировки, при котором два разных потока получают блокировки в несогласованном порядке.
public class Main {
private final Object lockA = new Object();
private final Object lockB = new Object();
public static void main(String[] args) {
new Main();
}
public Main() {
new Thread(new Runnable() {
public void run() {
a();
sleep(3000L); // Add a delay here to increase chance of deadlock.
b();
}
}, "Thread-A").start();
new Thread(new Runnable() {
public void run() {
// Note: Second thread acquires locks in the reverse order of the first!
b();
sleep(3000L); // Add a delay here to increase chance of deadlock.
a();
}
}, "Thread-A").start();
}
private void a() {
log("Trying to acquire lock A.");
synchronized(lockA) {
log("Acquired lock A.");
}
}
private void b() {
log("Trying to acquire lock B.");
synchronized(lockB) {
log("Acquired lock B.");
}
}
private void sleep(long millis) {
try {
Thread.sleep(millis);
} catch(InterruptedException ex) {
}
}
private void log(String msg) {
System.err.println(String.format("Thread: %s, Message: %s",
Thread.currentThread().getName(), msg));
}
}
Следующий код демонстрирует ситуацию, которая может привести к несовместимым результатамотсутствие контроля параллелизма между двумя потоками.
public class Main {
// Non-volatile integer "result".
private int i;
public static void main(String[] args) {
new Main();
}
public Main() {
Thread t1 = new Thread(new Runnable() {
public void run() {
countUp();
}
}, "Thread-1");
Thread t2 = new Thread(new Runnable() {
public void run() {
countDown();
}
}, "Thread-2");
t1.start();
t2.start();
// Wait for two threads to complete.
t1.join();
t2.join();
// Print out result. With correct concurrency control we expect the result to
// be 0. A non-zero result indicates incorrect use of concurrency. Also note
// that the result may vary between runs because of this.
System.err.println("i: " + i);
}
private void countUp() {
// Increment instance variable i 1000,000 times. The variable is not marked
// as volatile, nor is it accessed within a synchronized block and hence
// there is no guarantee that the value of i will be reconciled back to main
// memory following the increment.
for (int j=0; j<1000000; ++j) {
++i;
}
}
private void countDown() {
// Decrement instance variable i 1000,000 times. Same consistency problems
// as mentioned above.
for (int j=0; j<1000000; ++j) {
--i;
}
}
}