synchronization : Threads execute two critical sections in same order - java

I have the following kind of code:
synchronized block1 {
//only one thread in the block
}
{lot of code where synchronization not necessary}
synchronized block2 {
//only one thread in the block.
//All the threads that executed block1 before this thread should have already executed this block.
}
Each thread first executes block1, non synchronized block, and block2 in that same order.
If thread T1 executes block1 before thread T2, then T1 should execute block2 before T2. There are more than two threads.
Is there a way to achieve this in java?

As I understand Critical Section #2 MUST be executed in the same order as Critical Section #1
If thread T1 executes block1 before thread T2, then T1 should execute block2 before T2. There are more than two threads.
Then a Queue might be used to ensure the order of execution.
private Object lock = new Object();
private Queue<Thread> threadQueue = new ArrayDeque<>();
// https://stackoverflow.com/questions/32353283/synchronization-threads-execute-two-critical-sections-in-same-order
public void executeCriticalSectionsInOrder() throws InterruptedException {
// Critical Section #1
synchronized (lock){
// synchronized code #1
// Add self to queue
threadQueue.add(Thread.currentThread());
}
// {lot of code where synchronization not necessary}
// Critical Section #2
synchronized (lock) {
//All the threads that executed block1 before this thread should have already executed this block.
// Wait turn
Thread t = threadQueue.element(); // Do not remove until it is self
while (t != Thread.currentThread()) {
lock.wait();
// After sleep try again
t = threadQueue.element();
}
// Verified own turn. Update status
threadQueue.remove();
// synchronized code #2
lock.notifyAll(); // Awake any waiting thread after exiting section.
}
However If one thread dies/exits without removing itself from the queue, then following threads will be blocked indefinetely. Maybe add a finally block to do the housekeeping?
Note: In Nicholas Robinson's answer a position order was suggested instead of a queue, which seems slightly more efficient.

This basically creates a queue that threads will wait in until their number comes up. [UPDATED]
private AtomicInteger place = new AtomicInteger(0);
private AtomicInteger currentPlaceInQueue = new AtomicInteger(0);
private ReentrantLock lock = new ReentrantLock();
private Condition notNext = lock.newCondition();
public void method() {
ThreadLocal position = new ThreadLocal();
synchronized(this) {
//Your code
position.set(place.getAndIncrement());
}
// More code
lock.lock();
while ((int) currentPlaceInQueue.get() != position.get()) {
notNext.await();
}
// More code
lock.unlock();
currentPlaceInQueue.getAndIncrement();
notNext.notifyAll();
}

The synchronized blocks in your example are a red herring. Your problem is, you have N threads, and you have two blocks of code, and you want to make sure that none of the threads enters the second block until all of them have finished the first block.
You use a CyclicBarrier for that. http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/CyclicBarrier.html

You should be able to use a Lock which you take before calling block1 and release after calling block2.
static Lock lock = new ReentrantLock();
Random random = new Random();
public void block1() throws InterruptedException {
System.out.println("Enter block 1");
Thread.sleep(random.nextInt(500));
System.out.println("Leave block 1");
}
public void block2() throws InterruptedException {
System.out.println("Enter block 2");
Thread.sleep(random.nextInt(500));
System.out.println("Leave block 2");
}
private class BlockTester implements Runnable {
long start = System.currentTimeMillis();
#Override
public void run() {
while (System.currentTimeMillis() < start + 10000) {
lock.lock();
try {
System.out.println("Thread: " + Thread.currentThread().getName());
block1();
block2();
} catch (InterruptedException ex) {
System.out.println("Interrupted");
} finally {
lock.unlock();
}
}
}
}
public void test() throws InterruptedException {
Thread[] blockTesters = {
new Thread(new BlockTester()),
new Thread(new BlockTester()),
new Thread(new BlockTester()),
new Thread(new BlockTester()),
new Thread(new BlockTester())
};
for (Thread t : blockTesters) {
t.start();
}
for (Thread t : blockTesters) {
t.join();
}
}

Related

Why does this contrived Java code deadlock?

I'm having a hard time understanding synchronized. Since the first thread doesn't do anything with the object 2, doesn't it "unlock" everything in a second?
public class Uninterruptible {
public static void main(String[] args) throws InterruptedException {
final Object o1 = new Object(); final Object o2 = new Object();
Thread t1 = new Thread() {
public void run() {
try {
synchronized(o1) {
Thread.sleep(1000);
synchronized(o2) {}
}
} catch(InterruptedException e) { System.out.println("t1 interrupted"); }
}
};
Thread t2 = new Thread() {
public void run() {
try {
synchronized(o2) {
Thread.sleep(1000);
synchronized(o1) {}
}
} catch(InterruptedException e) { System.out.println("t2 interrupted"); }
}
};
t1.start(); t2.start();
Thread.sleep(2000);
t1.interrupt(); t2.interrupt();
t1.join(); t2.join();
System.out.println("Donezo!");
}
}
It doesn't matter that the inner synchronized blocks do nothing. Java will still attempt to acquire the lock on the the object specified.
No matter whether you have nothing or a huge amount of processing in the inner synchronized blocks, what you have is the minimal example to create a deadlock: two different threads, each owning the lock on a distinct resource, each attempting to acquire the lock on each other's resource.
The deadlock occurs before either thread even gets to executing the inner synchronized block, because neither thread can get locks on both resources at the same time.
The code winds up doing nothing but hanging, with each thread blocked. Your calls to interrupt are too late to cause an InterruptedException; they only set the "interrupted status" in the Thread. Commenting out Thread.sleep(2000) will let the calls to interrupt catch the Threads while they are still sleeping, before they even attempt to acquire the second lock.

For below program, when threads finish their work, why doesn't main thread stops?

This program works fine by printing alternate numbers via different threads but when all the numbers from 0-9 are printed Why does this program not stop? I have to manually stop my application.
public class EvenOddPrinter implements Runnable{
private AtomicInteger num = new AtomicInteger(0);
private Object lock = new Object();
#Override
public void run() {
synchronized (lock){
while (num.get()<10){
System.out.println(num.getAndAdd(1) + " - "+Thread.currentThread().getName());
lock.notify();
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
public class Executor {
public static void main(String[] args) throws InterruptedException {
EvenOddPrinter eop = new EvenOddPrinter();
Thread t1 = new Thread(eop);
Thread t2 = new Thread(eop);
t1.start();
t2.start();
}
}
that's because in the last Thread getting stuck at wait. notifyAll will notify all waiting thread if any there and release lock.
while (num.get()<10){
// existing implementation
}
lock.notifyAll();
The second thread t2 keeps waiting on the lock in the end, and t1 doesnt do the notify() anymore because the while condition becomes false. You must put a lock.notify(); statement outside of the while loop.
As soon as the number reached 8 the first thread calls notify() and goes to wait(). Second thread then makes the number 9 and calls notify() and goes to wait(). First thread is then not able to go inside the loop as specified in the condition, therefore, it exits the synchronized and block and finishes but second thread is still waiting. There has to be a mechanism to notifyAll() as soon as one of the threads exits the synchronized block which is exactly what I did.
#Override
public void run() {
synchronized (lock){
while (num.get()<10){
System.out.println(num.getAndAdd(1) + " - "+Thread.currentThread().getName());
lock.notify();
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notifyAll();
}
Also, lock is pointless when I am using AtomicInteger (or the other way round).

common data in two threads

I want to have two threads. One will increment my variable and the other one will show it on the screen just after it is incremented.
I want to use the wait() and notifyAll() functions to do that but I have some problems.
I have written this code, but it stops working at some point (it shows only the first and the last number: 1 and 10).
public class TestClass {
static int x = 0;
public static Object lock = new Object();
public static void main(String[] args) {
Thread t1 = new Thread(new Runnable() {
public void run() {
try {
for(int i = 0; i < 10; i++) {
synchronized (lock) {
lock.wait();
System.out.println(x);
}
}
}
catch (InterruptedException e) {
e.printStackTrace();
}
}
});
t1.start();
Thread t2 = new Thread(new Runnable() {
public void run() {
for(int i = 0; i < 10; i++) {
synchronized (lock) {
x++;
lock.notifyAll();
}
}
}
});
t2.start();
}
}
I have tried adding additional notifyAll() in the first thread and wait() in the second one(t2) but it still doesn't work. How can I do that right?
From JavaDoc:
The awakened threads will not be able to proceed until the current
thread relinquishes the lock on this object. The awakened threads will
compete in the usual manner with any other threads that might be
actively competing to synchronize on this object; for example, the
awakened threads enjoy no reliable privilege or disadvantage in being
the next thread to lock this object.
The thread that you just woke up is not guaranteed to acquire the lock immediately, thus you could have the loop that calls notifyAll() run all its iterations before the other gets woken up.
If you want them to alternate, one way is to have each one take turns waiting
Below is a working solution. Key is in adding an additional flag that will indicate whether the writing thread has the right to write. Note final on the lock object, that is a good practice.
public class TestClass
{
private static int x = 0;
private static final Object lock = new Object();
private static boolean canWrite = false;
public static void main(String[] args) {
Thread t1 = new Thread(new Runnable() {
public void run() {
try {
for(int i = 0; i < 10; i++)
{
synchronized(lock)
{
if(!canWrite)
lock.wait();
System.out.println(x);
canWrite = false;
lock.notify();
}
}
}
catch (InterruptedException e) {}
}
});
Thread t2 = new Thread(new Runnable() {
public void run() {
try {
for(int i = 0; i < 10; i++)
{
synchronized(lock)
{
x++;
canWrite = true;
lock.notify();
lock.wait();
}
}
} catch (InterruptedException ex) {}
}
});
t1.start();
t2.start();
}
}
The pattern is confusing but if you want to try to control threads, I suppose it's a useful exercise.
Change your code to give your thread a good name e.g. "test1" and "test2", say. Run your program and dig out the process ID for it and run
jstack <pid>
(you can find it in the bin directory of the JDK). This is (perhaps) what you will see:
test1 will be waiting one the
lock.wait();
while test2 will waiting on
lock.wait();
as well.
Start at the first iteration. test2 wins and calls notify, which has no effect at all because nothing is waiting on the lock. test2 then starts waiting.
Now test1 can enter and immediately goes into a wait.
If test1 wins the first time, you can get into a cycle that appears to work.
That is why you get the first number (test2 wins first) or the last number (test1 wins).
That's pretty much what it looks like to me, but your jstack will tell you for sure.
PS: Use CountDownLatch and the concurrency package in general.

Implementing deadlock condition

I am trying to implementing deadlock condition but somehow I am not able to get it working. Both the threads Thread1 and Thread2 are entering in the run function but only one of them enters in Sub/Sum depending on who entered run first. Example : if Thread2 entered run first the it will call sub() and Thread1 never calls sum(). I have also added sleep time so that Thread2 sleeps before calling sum() and Thread1 gets enough time to enter Sum() but Thread1 never enters.
public class ExploringThreads {
public static void main(String[] args) {
// TODO Auto-generated method stub
threadexample a1 = new threadexample();
Thread t1 = new Thread(a1, "Thread1");
Thread t2 = new Thread(a1,"Thread2");
t1.start();
t2.start();
}
}
class threadexample implements Runnable{
public int a = 10;
public void run(){
if(Thread.currentThread().getName().equals("Thread1"))
sum();
else if(Thread.currentThread().getName().equals("Thread2"))
sub();
}
public synchronized void sum()
{
try {
Thread.sleep(2000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(Thread.currentThread().getName()+"In Sum");
sub();
}
public synchronized void sub()
{
try {
Thread.sleep(2000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(Thread.currentThread().getName()+"In Sub");
sum();
}
}
If you really want to create an artificial dead lock, try this:
Thread1 and Thread2 are two threads that want to access the same file.
Thread1 starts, asks for a lock on File1.docx and sleeps for 2 minutes.
Thread2 starts, and makes an exclusive lock on File2.docx and now wants to access File1.docx.
Thread1 wakes up and now wants to access File2.docx which is held by Thread2
Now, this is a circular wait condition
Simple ? =)
This is not how you get a deadlock. Actually this code seems pretty safe :-) Only one thread enters sum/sub at a time because you are using synchronized which synchronizes on "this". There is only one "this" so both threads try to acquire the same lock.
Deadlock occurs, for instance, when Thread1 has one lock, Thread2 has second lock and then Thread1 would like to acquire Thread2's lock while still holding it's lock and Thread2 would like to acquire Thread1's lock while still holding it's lock.
What you could do is:
a) add 2 objects for locking in "threadexample" class (btw classes by convention should start with uppercase):
private final Object sumLock = new Object();
private final Object subLock = new Object();
b) drop the "synchronized" keyword in both sum/sub methods and instead use the synchronized() {} block in each of them. Sum would be synchronized(sumLock) { /* sum's body goes here / } and sub would be synchronized(subLock) { / sub's body goes here */}.
In this case Thread1 would go into sum(), acquire the sumLock and wait. Thread2 would go into sub(), acquire the subLock() and wait. Thread1 would wake up, go into sub() and try to acquire subLock but it's being held by Thread2 so it wait's until Thread2 releases it. In that time Thread2 wakes up, goes into sum() and tries to acquire sumLock which is held by Thread1 so Thread2 waits for Thread1 to release it.
Neither thread will go forward as each one of them is waiting for the other - you have a deadlock.
#Edit: yes you have only 1 instance of "threadexample" and both Thread1 and Thread2 are fighting for the lock but when one of them acquires the lock it will release it after executing sum/sub or sub/sum. For instance let's say Thread1 is first and starts executing sum(). It has the lock. In that case Thread2 will not go into sub() as it is protected by the same lock as Thread1. Thread1 will do sum(), then sub() and then it will release the lock --> Thread2 will go into sub() etc.
This is a working example of 'Deadlock in Action'. Basically what you need to do (and how that usually happens in real world) is that object are locked in opposite order: a first, b second in one thread and b first, a second in another:
package stackoverflow;
public class Deadlock {
final static String a = new String("A");
final static String b = new String("B");
public static void main(String[] args) {
final Thread abLock = new Thread() {
#Override
public void run() {
lock(a, b);
}
};
final Thread baLock = new Thread() {
#Override
public void run() {
lock(b, a);
}
};
abLock.start();
baLock.start();
}
static void lock(String first, String second) {
synchronized (first) {
System.out.println(first);
sleep();
synchronized (second) {
System.out.println(second);
}
}
}
static void sleep() {
try {
Thread.sleep(500);
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
}
}

Synchronization with threads

I have a two part question...
I have a class with a function in it that can only be accessed by any one thread at a given time. Making this a synchronized function or a synchronized block still allows for multiple threads since different threads are accessing it within the class. How can I make sure only one thread accesses this code? (See code example below)
With the synchronized function, the calls to the function are queued up. Is there any way to only allow the last call to the function to access the code? So if I have Thread1 currently accessing my function, then Thread2 and Thread3 try to access it (in that order) only Thread3 will be given access once Thread1 is complete.
public void doATask() {
// I create a new thread so the interface is not blocked
new Thread(new Runnable() {
#Override
public void run() {
doBackgroundTask();
}
}).start();
}
private void doBackgroundTask(MyObject obj) {
// perform long task here that is only being run by one thread
// and also only accepts the last queued thread
}
Thanks for any help!
If the second thread in your example can just return, you could use a combination of a lock and keeping track of the last thread executing the method. It could look like this:
private volatile Thread lastThread;
private final ReentrantLock lock = new ReentrantLock();
private void doBackgroundTask(Object obj) throws InterruptedException {
Thread currentThread = Thread.currentThread();
lastThread = currentThread;
try {
// wait until lock available
lock.lockInterruptibly();
// if a thread has arrived in the meantime, exit and release the lock
if (lastThread != currentThread) return;
// otherwise
// perform long task here that is only being run by one thread
// and also only accepts the last queued thread
} finally {
lock.unlock();
}
}
Full working test with additional logging that shows the thread interleaving and that T2 exits without doing nothing:
class Test {
private volatile Thread lastThread;
private final ReentrantLock lock = new ReentrantLock();
public static void main(String[] args) throws Exception {
final Test instance = new Test();
Runnable r = new Runnable() {
#Override
public void run() {
try {
instance.doBackgroundTask(null);
} catch (InterruptedException ignore) {}
}
};
Thread t1 = new Thread(r, "T1");
Thread t2 = new Thread(r, "T2");
Thread t3 = new Thread(r, "T3");
t1.start();
Thread.sleep(100);
t2.start();
Thread.sleep(100);
t3.start();
}
private void doBackgroundTask(Object obj) throws InterruptedException {
Thread currentThread = Thread.currentThread();
System.out.println("[" + currentThread.getName() + "] entering");
lastThread = currentThread;
try {
// wait until lock available
lock.lockInterruptibly();
// if a thread has arrived in the meantime, exit and release the lock
if (lastThread != currentThread) return;
// otherwise
// perform long task here that is only being run by one thread
// and also only accepts the last queued thread
System.out.println("[" + currentThread.getName() + "] Thinking deeply");
Thread.sleep(1000);
System.out.println("[" + currentThread.getName() + "] I'm done");
} finally {
lock.unlock();
System.out.println("[" + currentThread.getName() + "] exiting");
}
}
}
Output:
[T1] entering
[T1] Thinking deeply
[T2] entering
[T3] entering
[T1] I'm done
[T1] exiting
[T2] exiting
[T3] Thinking deeply
[T3] I'm done
[T3] exiting
What you want is probably a worker thread that waits for a signal to do some work. doATask() simply sends a signal to trigger the work. Accumulative signals are equivalent to one signal.
final Object lock = new Object();
MyObject param = null;
public void doATask(arg)
synchronized(lock)
param=arg;
lock.notify();
MyObject awaitTask()
synchronized(lock)
while(param==null)
lock.wait();
tmp=param;
param=null;
return tmp;
// worker thread
public void run()
while(true)
arg = awaitTask();
doBackgroundTask(arg);

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