If I use synchronize(this) in two methods and one calls the other, will I get stuck in a deadlock situation or will it work because the thread already owns the lock?
Picture the class below:
public class Test {
public void foo() {
synchronize(this) {
bar();
}
}
public void bar() {
synchronize(this) {
// do something
}
}
}
As you can see, there are two methods foo and bar, which both rely on synchronization.
When calling foo(), a lock will be obtained on (this); will bar try to do the same when called by foo (and thus causing a deadlock) or will it realize that the lock has already been obtained by the same thread?
Hope my explanation is more or less clear ;-)
The synchronized block is reentrant (in fact, Java monitors are reentrant, to be perfectly clear), thus no deadlock can happen in your situation.
According to the docs:
Recall that a thread cannot acquire a lock owned by another thread.
But a thread can acquire a lock that it already owns.
If thread holds lock of object it can enter to other synchronized blocks based on that lock object.
Here you can read that
"...thread can acquire a lock that it already owns. Allowing a thread to acquire the same lock more than once enables reentrant synchronization. This describes a situation where synchronized code, directly or indirectly, invokes a method that also contains synchronized code, and both sets of code use the same lock. Without reentrant synchronization, synchronized code would have to take many additional precautions to avoid having a thread cause itself to block."
One thing to be careful of though is if:
Thread A has the lock in foo() and needs to call bar()
and Thread B has the lock in bar() while needing to call foo()
Related
Are synchronized methods reentrant?
I have this code:
public class Main {
synchronized void m1() {
//some code
m2();
//some code
}
synchronized void m2(){
//some code
}
}
Say two threads (Thread A and Thread B) try to access m1() at the same time. Thread A takes lock first. After some time Thread A calls m2(), which is also synchronized on same object.
As I don't want to make any guesses can someone tell me this: Is there any guarantee what will happen when Thread A calls m2()? I mean once Thread A calls m2(), does it 'leave lock' at all? Because technically it left method's frame, so will it leave lock as well?
[EDIT]: Does Thread A get to run m2() every time once it calls it? Or does it leave lock so both Thread A and Thread B would race for their goals?
In Java locks are reentrant. If Thread A holds the lock, then it can start executing method m1. Eventually it will invoke method m2, since Thread A already holds the lock, it does not have to reacquire the lock again to execute method m2. Internally the lock keeps the owner field to track this. Even if the lock is already held, since the caller of m2 is the owner of the lock, the access is granted. This design strategy prevents deadlocks.
Assume the following class:
public class MyClass {
public synchronized void methodA() {
//...code
}
public synchronized void methodB() {
//...code
}
public synchronized void methodC() {
//...code
}
}
Assume that none of the synchronized methods of MyClass call each other.
Do I assume correctly that 1, and only 1 Thread may have access to ANY of the code of MyClass at any given time? If, for example, a thread is executing methodA() on an instance of MyClass, I assume that no other Thread may call either methodB() or methodC() at this time, but will block until the execution of methodA() by the first thread is complete.
I want to clarify my understanding of Goetz, Java Concurrency in Practice (28), who states:
"Acquiring the lock associated with an object does not prevent other threads from accessing that object--the only thing that acquiring a lock prevents any other thread from doing is acquiring that same lock."
In this particular case, I argue, Goetz' first statement is incorrect. MyClass employs implicit locking on each of its 3 methods--for each method the implicit lock is itself (this). Therefore, in this case, if a thread holds the implicit lock while executing methodA(), all other threads will be prevented from accessing any of the code on this object.
Do I understand method synchronization and implicit locking correctly?
Any perspectives are greatly appreciated!
"Acquiring the lock associated with an object does not prevent other
threads from accessing that object--the only thing that acquiring a
lock prevents any other thread from doing is acquiring that same
lock."
Your three methods are declared as being synchronized. When one of those methods is invoked, like
MyClass instance = new MyClass();
instance.methodA();
the currently executing thread (call it A) will acquire that instance's monitor. No other thread will be able to acquire the same object's monitor until A has released it. Nothing prevents other threads from calling methods on that same object.
Say you had thread A doing
instance.methodA();
and thread B doing
instance.methodB();
Then, yes, thread B would have to wait for methodA to complete (ie. for thread A to release the lock on the object referenced by instance).
However, if thread A was doing
instance.methodA();
and thread B was doing
instance.toString();
thread B would not have to wait at all since toString() does nothing with the object's monitor.
The point of this question is to illustrate that Java is not working as I expected.
How would you expect the following code to behave?
public class SynchTester {
private static SynchTester synchTester;
public synchronized static SynchTester getSynchTester(){
if(synchTester==null){
synchTester = new SynchTester();
}
return synchTester;
}
private SynchTester() {
SynchTester myTester = getSynchTester();
}
public static void main(String[] args) {
SynchTester tester = SynchTester.getSynchTester();
}
}
I would expect it to hang with a deadlock waiting on the recursion to complete, but instead it throws StackOverflow. Evidently synchronized does not block access to the same thread.
Is this a bug?
In Java, synchronized locks are reentrant.
Recall that a thread cannot acquire a lock owned by another thread. But a thread can acquire a lock that it already owns. Allowing a thread to acquire the same lock more than once enables reentrant synchronization. This describes a situation where synchronized code, directly or indirectly, invokes a method that also contains synchronized code, and both sets of code use the same lock. Without reentrant synchronization, synchronized code would have to take many additional precautions to avoid having a thread cause itself to block.
Source: see bottom of this page
A synchronized method needs to be able to get a lock on the monitor object. The monitor object is the instance (or class for a static method). A thread that already has the lock does not need to get it again. So yes, that could cause a stackoverflow (harhar).
from the java tutorials:
When one thread is executing a synchronized method for an object, all other threads that invoke synchronized methods for the same object block (suspend execution) until the first thread is done with the object.
So I think the syncronized keyword worked as expected, and a synchronized recursive call is perfectly legal (and working) in java.
Can you tell me if following invocations are reentrant or not?
public class Foo {
public synchronized void doSomething() {}
public synchronized void doAnotherSomething() {}
}
public class Too {
private Foo foo;
public synchronized void doToo() {
foo.doSomething();
//...other thread interfere here....
foo.doAnotherSomething();
}
}
are 2 continuous invocations in method doToo() reentrant? I'm not sure about this case since foo.doSomething() method acquire and release the intrinsic lock, no nested synchronization between 2 invocations. Is there situation that other thread might interfere between 2 invocations?
First of all, regarding reentrant locks in Java:
Synchronized blocks in Java are reentrant. This means, that if a Java thread enters a
synchronized block of code, and thereby take the lock on the monitor object the block is synchronized on, the thread can enter other Java code blocks synchronized on the same monitor object.
Taken from here.
The two consecutive calls you described (in doToo) will not be interfered unless another object has a reference to Too's private Foo, since, to access foo, one needs to lock Too. However, the calls do not invoke reentry as the locks are acquired and released for every call. They would be reentrant if doSomething called doAnotherSomething or vice versa.
That depends entirely on what other threads are accessing. Can another thread take over the CPU between those functions? Yes. Will there be a race condition? Depends on many factors.
From what you posted foo.doSomething will be locked on foo, then released, then locked again upon entry to doAnotherSomething. So if another thread not locked on the same Too object tries to manipulate foo they will be able to do so between doSomething and doAnotherSomething. If everyone synchronized on the same Too object before manipulating the underlying foo object, then those two methods of foo will not have state manipulated between calls because the Too object method will block other threads until completion. Thus if you have a problem or not depends on your other accessors.
There is no reentrance here from what you posted, but java is ok with reentrant synchronization as Amir posted.
first of all locks are applicable on objects so need to create the object and then apply the lock.
are 2 continuous invocations in method doToo() reentrant? I
In your case they are not re-entrant . if code is something like below then they will be re-entrant .
public class Foo {
public synchronized void doSomething() {
doAnotherSomething();
}
public synchronized void doAnotherSomething() {}
}
Once lock has been acquired on one object then on same object it can traverse like in above case.
Assume the following class
public class TestObject{
public void synchronized method1(){
//some 1000 lines of code
}
public void method2(){
//some 1000 lines of code
}
}
Let's assume there are two threads accessing the same instance of TestObject class, let's call them t1 and t2. I want to know what will happen in the following scenario.
When t1 is in midway of accessing method1(). Now t2 is trying to access method2().
When t1 is in midway of accessing method2(). Now t2 is trying to access method1().
My understanding is that for the first question, the thread t2 will not be given permission as the object will be locked by t1. For the second question, the thread t2 will be granted access and takes lock on the object and will stall t1 from execution. But my assumption was wrong. Can anyone explain this?
Thanks
Only the method with the keyword synchronized holds a lock for this object when a thread is running in that method.
Had both method 1 and method 2 been declared as synchronized, one thread would block the other even though they are trying to run different methods.
In your example only 1 method is blocking by an implicit lock.
As a result t1 and t2 can be running concurrently in method 1 and method 2 (or vice versa).
Only when trying to access method 1, a t1 or t2 would block if the lock has already been acquired
When you declare a method to be synchronized, e.g.:
public synchronized void foo() {
// Do something
}
the compiler treats it as though you had written this:
public void foo() {
synchronized (this) {
// Do something
}
}
In your example you have one synchronized method and one non-synchronized. This means that only access to method1 will be locked. Locking checks are only done on entry to a synchronized block, so calling method2 will not trigger any locking.
To answer your two questions, then, in both cases the two threads will be allowed to proceed because they are not trying to obtain a lock on the same object. If you declare method2 to be synchronized (or manually add a synchronized (this) block) then one thread will be forced to wait for the other.
Remember: synchronizing on an object does not prevent other threads calling methods on that object. It only prevents another thread entering a synchronized block with the same lock object.
Incidentally, it's often better to have an internal lock object rather than declaring methods to be synchronized, e.g.
class Foo {
private final Object LOCK = new Object();
public void doSomething() {
synchronized (LOCK) {
// Whatever
}
}
}
Otherwise I can break your thread-safety by doing this:
class MessEverythingUp {
public MessEverythingUp(Foo foo) {
synchronized (foo) {
while (true) {
System.out.println("Pwnd ur thread safety");
}
}
}
}
Since I'm locking the instance of foo, your synchronized methods (with their implicit "synchronized (this)") will not be able to obtain a lock and will block forever. Most importantly, you cannot prevent this because I can synchronize on whatever object I like. Obviously the example is extreme but you can get nasty, subtle deadlock bugs if you're not careful about this sort of thing.
In both cases, the second thread will be given permission to execute its method.
Since only one of these two methods contains the synchronized keyword, both of these methods can be executed simultaneously. The restriction is that only one method with that keyword can be executed at any given time, because executing that method requires a lock on the object. A thread without the keyword requires no lock and will always be allowed to execute regardless of the object being locked.
I also assume here that the 1000 lines of code in method2 does not contain a block like this:
synchronized (this) {
}
If it does, then the results will be different.
t2 will be allowed to start executing the method. When it hits the synchronized line, it will wait for t1 to finish method1 and release its lock before continuing.
t2 will be allowed to start executing the method so long as t1 is not within the synchronized code block. If it is, t2 will wait until t1 exits the block and releases the lock before beginning. If t1 has not yet entered the synchronized block, then t2 will acquire the lock, and t1 will wait as soon as it gets to the synchronized code block until t2 completes the method and releases the lock.
One method is synchronized while the other is not. So no matter whether the lock on an Object (in this is case the instance the method belongs to) has been acquired or not, the non-synchronized method will execute unimpeded (since it foes not try to acquire or wait for a lock). Which means in both cases both threads will run without waiting for each other - resulting in a possibly inconsistent state of the object.
method1() is synchronized and hence called as thread safe method. When multiple threads try to access this method simultaneously then only the lock on instance object will work.
method2() is not synchronized and hence is a thread unsafe method, other threads can call this method even if the some other thread has lock on the instance, that is why this method is called as thread unsafe method.
In both cases you mentioned one thread will get lock on the instance by calling method1() and other thread will try to access method2() which is unsafe and hence both thread will execute.
with regards
Tushar Joshi, Nagpur
Both threads will execute as if the lock does not exist.
The thread accessing method2 will never know that it should get a lock so it will execute even if an other thread holds the lock. Only synchronized methods will reserve the lock since synchronization is optional and not always wanted or even necessary.
If both threads execute method1 one of them will block until the other thread exits the method and releases the lock.
To make sure only one thread executes and all others wait you have to make both methods synchronized.