I was just curious is it possible that a thread T1 say executes a synchronization block partially and then releases the lock on the object and another thread T2 executes the same block? Something like this:
line1: synchronized(this){
line2: if(INSTANCE == null)
line3: INSTANCE = new Object(); //say a variable is initialized
line4: return INSTANCE;
line5: }
Is it possible that thread T1 acquires a lock on current object (this) and executes line1 and line2. Then thread T1 is preempted by thread T2, T1 releases lock and T2 acquires lock on this and executes the same block (all the lines1 to 5). Then thread T1 again takes the lock and continues executing from line3?
Basically, T1 will see INSTANCE as null and so will T2 and each will create a new Object.
If this is not possible can someone explain why not?
Addendum:
Thanks everyone for your answer. My question was a bit misleading. What I am exactly asking, is it possible that once a thread is executing a synchronized block it can release the lock before the entire block is executed (not by explicitly calling wait() but something which is process, CPU dependent)? Is there a contract in JLS or a JVM guarantee that once a thread starts executing a synchronized block the lock on the object is not released until the end of the block? My understanding is synchronization guarantees no 2 threads can simultaneously execute the block (or other synchronized method/block) but the lock is hold until the end of the block is reached? It's kind of obvious but is it specified in the JLS?
Thread preemption doesn't cause the preempted thread to release its locks. If it did, locks would be worthless. The whole point of a synchronized block is that it will disallow other threads from synchronizing on the same object until the thread releases the lock (by leaving the synchronized block).
Even if a thread is preempted, it won't release a lock. The lock is still held. If another thread comes along, it will block (stop running) until the lock is released, even if the original thread gets preempted several times before it releases the lock. Basically almost any kind of lock has some storage in the heap that gets written to indicate there's a lock. It's permanent until the thread or the system writes a different value to indicate the lock is free.
It is of course possible to write code that allows access to an instance or field without ever taking the lock, but that's a coding error. It's also possible for the original thread to exit the block early (say it throws an exception) -- this releases the lock, and other threads can continue as normal.
I'm pretty sure it's not possible for a second thread to enter the synchronize block before the first one has executed the entire block. On obtaining the lock on the object, all other threads attempting to enter the synchronized code will be blocked.
See more information here: http://tutorials.jenkov.com/java-concurrency/synchronized.html
Basically you can use Locks objects. Lock objects can allow you to sequentially apply and release locks on multiple locks line by line.
A very good tutorial on how to implement it here
Check out below code:
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
public class test2{
private static Object instance = null;
private static test2 t = new test2();
private static Lock lock = new ReentrantLock();
public static void main(String[] args) {
A a = new A();
B b = new B();
a.setName("Thread-A");
b.setName("Thread-B");
a.start();
b.start();
}
public Object getObj(){
try {
lock.lock();
System.out.println("Begin Current thread: "+ Thread.currentThread().getName());
if(instance == null){
if(Thread.currentThread().getName().equalsIgnoreCase("Thread-A")){
lock.unlock();
while(instance==null){
System.out.println(Thread.currentThread().getName() +" : instance is null");
}
while(true){
if(!lock.tryLock()){
System.out.println(Thread.currentThread().getName() + " waiting for re lock");
}else{
lock.lock();
break;
}
}
}
instance =new Object();
System.out.println("End Current thread: "+ Thread.currentThread().getName());
if (((ReentrantLock) lock).isHeldByCurrentThread()) {
lock.unlock();
}
}
} catch (Exception e) {
e.printStackTrace();
}
return instance;
}
public static class A extends Thread{
#Override
public void run() {
while(true){
if(t.getObj() != null){
break;
}
}
}
}
public static class B extends Thread{
#Override
public void run() {
while(true){
if(t.getObj() != null){
break;
}
}
}
}
}
Output
Begin Current thread: Thread-A
Thread-A : instance is null
Begin Current thread: Thread-B
Thread-A : instance is null
End Current thread: Thread-B
Thread-A waiting for re lock
End Current thread: Thread-A
Related
I came across the following e example to implement custom suspend and wait from some website.
// Suspending and resuming a thread the modern way.
class NewThread implements Runnable {
String name; // name of thread
Thread t;
boolean suspendFlag;
NewThread(String threadname) {
name = threadname;
t = new Thread(this, name);
System.out.println("New thread: " + t);
suspendFlag = false;
t.start(); // Start the thread
}
// This is the entry point for thread.
public void run() {
try {
for (int i = 15; i > 0; i--) {
System.out.println(name + ": " + i);
Thread.sleep(200);
synchronized(this) {
while (suspendFlag) {
wait();
}
}
}
} catch (InterruptedException e) {
System.out.println(name + " interrupted.");
}
System.out.println(name + " exiting.");
}
void mysuspend() {
suspendFlag = true;
}
synchronized void myresume() {
suspendFlag = false;
notify();
}
}
class SuspendResume {
public static void main(String args[]) {
NewThread ob1 = new NewThread("One");
NewThread ob2 = new NewThread("Two");
try {
Thread.sleep(1000);
ob1.mysuspend();
System.out.println("Suspending thread One");
Thread.sleep(1000);
ob1.myresume();
...................
I am more concerned about the ob1.mysuspend() and ob1.myresume() calls. When my suspend is called then ob1 will be placed into the blocking queue associated with the runnable object it is using. When ob1 calls myresume, then how does it work as ob1 is already in waiting queue for the same object, can the waiting object enters another synchronised method and then signals notify to itself?How does this work?What am I missing?
The thread is written so that while an instance of NewThread is running, another thread can call mysuspend to suspend that running thread. Again, a thread other than the suspended thread calls myresume to resume the suspended thread.
There also appears to be a data race because mysuspend writes to suspendFlag without any synchronization. That means, the thread that needs to be suspended may not see that write immediately. mysuspend must be declared synchronized, or suspendFlag must be volatile.
This code is flat out broken.
Straight up broken: JMM violation
The mysuspend method (which should be named mySuspend, by the way) updates a field that is then read from another thread, and isn't synchronized. This is an error - and a really nasty one because you cannot reliably test that it is an error. The Java Memory Model (JMM) states that any write to a field may be observable or not, at the discretion of the JVM implementation, unless a so-called Happens-Before/Happens-After relationship is established (there are many ways to do it; usually you do so via synchronized, volatile, or some other concurrency tool built on these primitives, such as the latches and queues in the java.util.concurrent package).
You do not establish such a relationship here, meaning, that suspendFlag = true results in a schroedingers cat variable: The other thread that reads this field may read true or false, the JVM gets to decide what you see. Hence: A bug, and, untestable. bad. Any field that is read/written to by multiple threads needs to be written extremely carefully.
Mark that method synchronized, that's a good first step.
Wait and Notify
You've got it flipped around: You must in fact hold the synchronized lock on x when you invoke wait on x (here, x is this).
To call x.wait() (you are calling this.wait(), effectively), you must first be in a synchronized(x) block. Once the wait 'goes through', the code releases the lock (other synchronized(x) blocks can run). To invoke x.notify() you must also hold that lock.
wait does not return until the lock is re-established.
In other words:
public void foo() {
wait();
}
will fail at runtime. Try it. Guaranteed exception. In the mean time, this:
public void foo() {
synchronized (this) {
// code before wait
wait();
// code after wait
}
}
is executed as if it is written like this:
public void foo() {
synchronized (this) {
// code before wait
release_lock(this);
this.wait();
acquire_lock(this);
// code after wait
}
}
Where acquire_lock is guaranteed to actually take a while (because by definition whatever invoked notify() to wake you up is currently holding it! So wait is always a two-hit thing: You need to be both notified AND the lock needs to be reacquired before your code will continue). Except, of course, acquire_lock and release_lock don't exist, and unlike this hypothetical code, wait() is more atomic than that.
I want to know if it's authorized to avoid Thread deadlocks by making the threads not starting at the same time? Is there an other way to avoid the deadlocks in the following code?
Thanks in advance!
public class ThreadDeadlocks {
public static Object Lock1 = new Object();
public static Object Lock2 = new Object();
public static void main(String args[]) {
ThreadDemo1 t1 = new ThreadDemo1();
ThreadDemo2 t2 = new ThreadDemo2();
t1.start();
try {
Thread.sleep(100);
} catch (InterruptedException e) {
}
t2.start();
}
private static class ThreadDemo1 extends Thread {
public void run() {
synchronized (Lock1) {
System.out.println("Thread 1: Holding lock 1...");
try {
Thread.sleep(10);
} catch (InterruptedException e) {
}
System.out.println("Thread 1: Waiting for lock 2...");
synchronized (Lock2) {
System.out.println("Thread 1: Holding lock 1 & 2...");
}
}
}
}
private static class ThreadDemo2 extends Thread {
public void run() {
synchronized (Lock2) {
System.out.println("Thread 2: Holding lock 2...");
try {
Thread.sleep(10);
} catch (InterruptedException e) {
}
System.out.println("Thread 2: Waiting for lock 1...");
synchronized (Lock1) {
System.out.println("Thread 2: Holding lock 1 & 2...");
}
}
}
}
}
There are two ways to get a deadlock:
Lock escalation. For example, a thread holding a shareable read
lock tries to escalate to an exclusive write lock. If more than one
thread holding a read lock tries to escalate to a write lock, a
deadlock results. This doesn't apply to what you're doing. (Offhand, I don't even know if it's possible to escalate a lock in Java.)
Unspecified lock order. If thread A locks object 1, then tries to lock object 2, while thread B locks object 2 then tries to lock object 1, a deadlock can result. This is exactly what you're doing.
Those are the only ways to get a deadlock. Every deadlock scenario will come down to one of those.
If you don't want deadlocks, don't do either of those. Never escalate a lock, and always specify lock order.
Those are the only ways to prevent deadlocks. Monkeying around with thread timing by delaying things is not guaranteed to work.
As the other mentioned, delays won't help because threads by their nature have unknown start time. When you call start() on a thread, it becomes runnable, but you cannot know when it will be running.
I'm assuming this is just demo code, so you already know that playing with sleeps is not guaranteed to work (as stressed in other answers).
In your demo code I see two options to try avoid the deadlock:
Remove any sleep within the body of the functions executed by the threads and just put a single, long enough, sleep between the start of the two threads; in practical terms, this should give enough time to the first thread to be scheduled and complete its work, then the second thread will acquire both locks without contention. But, you already know, scheduling policies are not under your control and this is not guaranteed to work at all.
Do acquire locks in the same order in both threads, without using any sleep at all, i.e.
synchronized (Lock1) {
synchronized (Lock2) {
// ...
}
}
This is guaranteed to remove any possible deadlock, because the first thread to acquire Lock1 will gain the possibility to complete its work while blocking the other thread until completion.
UPDATE:
To understand why acquiring locks in the same order is the only guaranteed way to avoid deadlock, you should recall what's the whole purpose of locks.
A thread is said to own a lock between the time it has acquired the lock and released the lock. As long as a thread owns a lock, no other thread can acquire the same lock. In fact, the other thread will block when it attempts to acquire the same lock.
Every object in Java has an intrinsic lock associated with it. The synchronized statement let you automatically acquire the intrinsic lock of the specified object and release it after code execution.
No, starting threads at different times is not a way to avoid deadlocks - in fact, what you'd be trying with different start times is a heuristic to serialize their critical sections. ++ see why at the and of this answer
[Edited with a solution]
Is there an other way to avoid the deadlocks in the following code?
The simplest way is to acquire the locks in the same order on both threads
synchronized(Lock1) {
// do some work
synchronized(Lock2) {
// do some other work and commit (make changes visible)
}
}
If the logic of your code dictates you can't do that, then use java.util.concurrent.locks classes. For example
ReentrantLock Lock1=new ReentrantLock();
ReentrantLock Lock2=new ReentrantLock();
private static class ThreadDemo1 extends Thread {
public void run() {
while(true) {
Lock1.lock(); // will block until available
System.out.println("Thread 1: Holding lock 1...");
try {
// Do some preliminary work here, but do not "commit" yet
Thread.sleep(10);
} catch (InterruptedException e) {
}
System.out.println("Thread 1: Waiting for lock 2...");
if(!Lock2.tryLock(30, TimeUnit.MILLISECOND)) {
System.out.println("Thread 1: not getting a hold on lock 2...");
// altruistic behaviour: if I can't do it, let others
// do their work with Lock1, I'll try later
System.out.println("Thread 1: release lock 1 and wait a bit");
Lock1.unlock();
Thread.sleep(30);
System.out.println("Thread 1: Discarding the work done before, will retry getting lock 1");
}
else {
System.out.println("Thread 1: got a hold on lock 2...");
break;
}
}
// if we got here, we know we are holding both locks
System.out.println("Thread 1: both locks available, complete the work");
// work...
Lock2.unlock(); // release the locks in the reverse...
Lock1.unlock(); // ... order of acquisition
}
}
// do the same for the second thread
++ To demonstrate why delays in starting the threads at different times is not a foolproof solution, think if you can afford to delay one of the threads by 10 seconds in the example below. Then think what will you do if you don't actually know how long to wait.
private static class ThreadDemo1 extends Thread {
public void run() {
synchronized (Lock1) {
System.out.println("Thread 1: Holding lock 1...");
try {
// modelling a workload here:
// can take anywhere up to 10 seconds
Thread.sleep((long)(Math.random()*10000));
} catch (InterruptedException e) {
}
System.out.println("Thread 1: Waiting for lock 2...");
synchronized (Lock2) {
System.out.println("Thread 1: Holding lock 1 & 2...");
}
}
}
}
private static class ThreadDemo2 extends Thread {
public void run() {
synchronized (Lock2) {
System.out.println("Thread 2: Holding lock 2...");
try {
// modelling a workload here:
// can take anywhere up to 10 seconds
Thread.sleep((long)(Math.random()*10000));
} catch (InterruptedException e) {
}
System.out.println("Thread 2: Waiting for lock 1...");
synchronized (Lock1) {
System.out.println("Thread 2: Holding lock 1 & 2...");
}
}
}
}
This question was asked to me in an interview. Before I had told him this,
Once a thread enters any synchronized method on an instance, no other
thread can enter any other synchronized method on the same instance.
Consider the snippet:
Q1:
public class Q1 {
int n;
boolean valueSet = false;
synchronized int get() {
while (!valueSet)
try {
wait();
} catch (InterruptedException e) {
System.out.println("InterruptedException caught");
}
System.out.println("Got: " + n);
valueSet = false;
notify();
return n;
}
synchronized void put(int n) {
while (valueSet)
try {
wait();
} catch (InterruptedException e) {
System.out.println("InterruptedException caught");
}
this.n = n;
valueSet = true;
System.out.println("Put: " + n);
notify();
}
}
Producer1:
public class Producer1 implements Runnable {
Q1 q;
Producer1(Q1 q) {
this.q = q;
new Thread(this, "Producer").start();
}
#Override
public void run() {
int i = 0;
while (true) {
q.put(i++);
}
}
}
Consumer1
public class Consumer1 implements Runnable {
Q1 q;
Consumer1(Q1 q) {
this.q = q;
new Thread(this, "Consumer").start();
}
#Override
public void run() {
while (true) {
q.get();
}
}
}
PC1:
public class PC1 {
public static void main(String args[]) {
Q1 q = new Q1();
new Producer1(q);
new Consumer1(q);
System.out.println("Press Control-C to stop.");
}
}
So, he asked as soon as you have created this thread new Producer1(q), then according to you, the synchronized int get() method must have been locked by the same thread, i.e, by new Producer1(q) when it accessed synchronized int put(). I said yes.
But I checked in eclipse, get is callable by new Consumer1(q). The program works perfect.
Where am I going wrong?
O/P:
The call to wait() will release the monitor for the time waiting.
That's what is documented for Object.wait():
The current thread must own this object's monitor. The thread
releases ownership of this monitor and waits until another thread
notifies threads waiting on this object's monitor to wake up
either through a call to the notify method or the
notifyAll method. The thread then waits until it can
re-obtain ownership of the monitor and resumes execution.
Once a thread enters any synchronized method on an instance, no other
thread can enter any other synchronized method on the same instance.
What you forgot to add here is "except if the lock is released".
...and it is the case in your example, when calling wait.
The documentation specify :
The thread releases ownership of this monitor and waits until another
thread notifies threads waiting on this object's monitor to wake up
either through a call to the notify method or the notifyAll method.
Since the lock is released, you step in the other method (and the condition is true because the boolean was modified). Once in the other method, you release the lock again, then call notify and you wake up the old thread which terminates (re-modify boolean to pass the condition in other method, and notify). That way you step between both methods ad-infinitum.
wait() and notify() is acts as a signal between threads, to control the threads to do or to not do the stuff.
The program works perfect because here 2 threads (Producer, Consumer) which fight for the one lock (monitor). When Consumer aquires the lock (Q1 object) then Producer is waiting for the lock. When Consumer finishes his work it release the lock. Consumer releases the lock also when wait() method has been called, because wait() sets thread to Waiting state with lock release. It's time for Producer to aquire the lock and does his work. When Producer thread notify() calls then Consumer continue his work (when aquired the lock). The same is right for Producer.
Resume: Q1 object is a lock for all threads. If it aquired someone then others are blocked and the answer is - it not possible to get an access at the same time to the get(), put() methods more then 2 threads.
I think that the question is ambiguous. (E.g., what does "accessible" mean?)
IMO, a good interview question should not have a right answer and a wrong answer. A good interview question should be a conversation starter, that gives you an opportunity to show how much you know about the subject.
When I am asking the interview questions, I like a candidate who can see through the question, and get down to the underlying mechanism. E.g.,
What the JLS guarantees is that no two threads can be _synchronized_
on the same instance at the same time...
Then we could explore questions like, how could two threads enter the same synchronized method at the same time? (e.g., synchronized on different instances), how could two threads be in the same synchronized method for the same instance at the same time (one of them could be in a wait() call), ...
A thread can not access a synchronized block of code unless it has aquired a lock on the object that guards the block. In your case, the synchronized keyword uses the lock of the object in which it has been declared. So as long as a thread is executing get(), no other thread can execute the put().
If you apply this, when put() sets the value, it notifies consumer which accepts the value. The code should work even after you have removed the wait() and notify() calls from both get and put methods
I write the below code to test when will the thread is awake when it is waiting for a Condition object.
But I find I have to unlock after I call signal(). Lock is not release by this method, while await() will release this lock .
This is from Condition#await
The lock associated with this Condition is atomically released and the current thread becomes disabled for thread scheduling purposes and lies dormant until one of four things happens:
And this is from Conditon#signal
Wakes up one waiting thread.
If any threads are waiting on this condition then one is selected for waking up. That thread must then re-acquire the lock before
returning from await.
But in my code, this is not true, until we unlock the lock. Why it is design like this? Since in my opinion, when we decide to to signal the others, we should not hold the lock any more,am I wrong?
Since we can do many things between calling signal and unlock ,say I sleep 10 seconds, what exactly the time java signal the other thread? Is there a another background thread who is working between we signal and unlock?
public class WorkerThread extends Thread{
#Override
public void run() {
Monitor.lock.lock();
while (!Monitor.isConditionTrue){
try {
Monitor.condition.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("THREAD ID "+this.getId()+"-------working --------");
System.out.println("------singnall--------");
Monitor.isConditionTrue=true;
Monitor.condition.signal();
try {
Thread.sleep(3000);//here, the thread is sleeping while another thread is not awaken since the lock is not releases
System.out.println("------unlock--------");
Monitor.lock.unlock();//now the other thread is awaken, if I do not explicitly unlock , no thread will be awaken.
}
catch (InterruptedException e) {
e.printStackTrace();
}
}
}
public class Monitor {
static ReentrantLock lock = new ReentrantLock();
static Condition condition = lock.newCondition();
static volatile boolean isConditionTrue = true;
public static void main(String args[]) {
Thread t1 = new WorkerThread();
Thread t2 = new WorkerThread();
t1.start();
t2.start();
Thread.sleep(2000);
lock.lock();
isConditionTrue=true;
condition.signalAll();
lock.unlock();
}
}
OUTPUT:
THREAD ID 9-------working --------
------singnall--------
------unlock--------
THREAD ID 10-------working --------
------singnall--------
------unlock--------
You have missed this sentence in Contition#await:
In all cases, before this method can return the current thread must re-acquire the lock associated with this condition. When the thread returns it is guaranteed to hold this lock.
In other words, you must explicitly release the lock after await, just as with signal.
Why this mechanism is sound: if you first released the lock, then signaled, you'd be open to race conditions where other threads made changes between releasing the lock and the signal reaching a parked thread. The way the mechanism works, first a definite thread is chosen to be awoken by the signal, then it waits for the lock, then the signaling thread releases it, then the awoken thread goes on.
You might argue that signal could do all of this internally, but then:
the API would become confusing: there would be more than one method releasing the lock;
the APi would become more restrictive and preclude any use cases where the thread wants to do something more before releasing the lock, such as atomically issuing more signals.
I wrote this program to check if a thread t1 holding lock on two different objects :
Lock.class and MyThread.class goes into waiting mode on MyThread.class instance using MyThread.class.wait().It does not release lock on Lock.class instance. why so ? I have been thinking that once a thread goes into wait mode or it dies it releases all the acquired locks.
public class Lock {
protected static volatile boolean STOP = true;
public static void main(String[] args) throws InterruptedException {
MyThread myThread = new MyThread();
Thread t1 = new Thread(myThread);
t1.start();
while(STOP){
}
System.out.println("After while loop");
/*
*
*/
Thread.sleep(1000*60*2);
/*
* Main thread should be Blocked.
*/
System.out.println("now calling Check()-> perhaps i would be blocked. t1 is holding lock on class instance.");
check();
}
public static synchronized void check(){
System.out.println("inside Lock.check()");
String threadName = Thread.currentThread().getName();
System.out.println("inside Lock.Check() method : CurrrentThreadName : "+ threadName);
}
}
class MyThread implements Runnable{
public MyThread() {
}
#Override
public void run() {
try {
System.out.println("inside Mythread's run()");
classLocking();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
public static synchronized void classLocking() throws InterruptedException{
System.out.println("inside Mythread.classLocking()");
String threadName = Thread.currentThread().getName();
System.out.println("inside MyThread.classLocking() : CurrrentThreadName : "+ threadName);
/*
* outer class locking
*/
synchronized (Lock.class) {
System.out.println("I got lock on Lock.class definition");
Lock.STOP = false;
/*
* Outer class lock is not released. Lock on MyThread.class instance is released.
*/
MyThread.class.wait();
}
}
}
You are correct that it doesn't release the other lock. As for why, it's because it isn't safe to do so. If it was safe to release the outer lock during the call to the inner function, why would the inner function be called with the other lock held at all?
Having a function release a lock it didn't acquire behind the programmer's back would destroy the logic of synchronized functions.
Yes it is working correctly. A thread goes into waiting status releases the corresponding lock instead of all locks. Otherwise think about that: if things are like what you thought, then when a thread waits it loses all the acquired locks, which makes advanced sequential execution impossible.
The semantics of wait() is that the Thread invoking it notices that a lock was already acquired by another thread, gets suspended and waits to be notified by the thread holding the lock when the latter one releases it (and invokes notify). It doesn't mean that while waiting it releases all the locks acquired. You can see the wait's invocations as a number of barriers the thread meets on the way to acquiring all the locks it needs to accomplish an action.
Regarding the question "Why a thread doesn't release all the locks acquired when invoking wait" , I think the answer is that, doing so would make it more prone to starvation and it would also slow down the progress in a multithreaded application (All threads would give up all their locks when invoking the first wait and would have to start over when they acquire the lock they are currently waiting for. So, they would be in a permanent battle for locks.
Actually, in such a system, the only thread able to finish execution would be the one which manages to find all locks free when it needs them. This is unlikely to happen)
From JavaDoc of method wait()
The current thread must own this object's monitor. The thread releases ownership of this monitor and waits until another thread notifies threads waiting on this object's monitor to wake up either through a call to the notify method or the notifyAll method. The thread then waits until it can re-obtain ownership of the monitor and resumes execution.