Let's say we have an object whose methods/fields are synchronized on "this". This question is actually about "this" as I think I have a hard time what "this" reference means.
So our object is:
class A {
private Field a;
private Field b;
public synchronized void doSomething() {
//something with a
}
public synchronized void somethingElse() {
//do something as with b
}
}
Then we have another object or method that takes A objects and do work on a and b via doSomething and somethingElse methods. So I need to keep the state consistent while I process A objects, hence I synchronize. Let's say those A objects are values of a Map. Then I am iterating over the values and do the things that I do. So the question is, is it thread-safe to do it in the following way:
for(A aObject : map.values()) {
synchronized(aObject) {
aObject.doSomething();
aObject.somethingElse();
}
}
If the "this" reference is the same reference as aObject, I think I shouldn't be in trouble. But what if I do it like this:
for(A aObject : map.values()) {
A anotherReference = aObject;
synchronized(anotherReference) {
anotherReference.doSomething();
anotherReference.somethingElse();
}
}
Is it still thread-safe? I mean can I synchronize on a local copy of a lock reference?
Note: this is an oversimplification of something I need to do in my code.
The synchronisation monitor belongs to the object that is referenced, not the reference, so your two for loops are equivalent, they both synchronise on the same object.
Now a synchronized method
public synchronized void foo() {
// do stuff
}
is exactly equivalent to
public void foo() {
synchronized(this) {
// do stuff
}
}
so in the loop
for(A aObject : map.values()) {
synchronized(aObject) {
aObject.doSomething();
aObject.somethingElse();
}
}
the synchronized block is locking the same monitor as the doSomething() and doSomethingElse() methods use themselves. What you gain from the synchronized block is that no other thread can sneak in and call either of those methods on the same A instance in between these two calls.
you seems to be confused about what references are, so i would go read up on them. when you use a synchronized block, you are not synchronizing on the reference itself, but on the object instance to which the reference refers.
for example:
Object a = new Object();
Object b = a;
synchronized(a) { ... }
synchronized(b) { ... }
these two synchronized blocks are synchronizing on the same Object instance because a and b refer to the same Object instance.
Following from that, a synchronized method is the same as synchronizing on the this reference.
for example:
public class A {
public synchronized void doStomething() { ... }
public void doSomethingElse() { synchronized(this) { ... } }
}
both of these methods are synchronizing on the same Object instance (the current instance), using the self reference known as this. you can rewrite either example as the other example and they are equivalent.
So, returning to your original example, i hope you will understand that when you synchronized on an Object instance externally through a reference (as my first example), it is doing the same thing as an Object synchronizing internally on itself.
To wrap, your last example is a common idiom when working on synchronized collections as it enables the caller to ensure that 2 operations are performed atomically with respect to the Collection.
for example:
// this will result in a List where all methods are internally synchronized
List<Object> syncList = Collections.synchronizedList(new ArrayList<Object>());
// i can safely perform an atomic operation on the List using this pattern
synchronized(syncList) {
if(syncList.isEmpty()) { // <- synchronized method call
syncList.add(...); // <- synchronized method call
}
}
In Java we have two basic synchronization idioms: synchronized methods and synchronized statements.
When you use the first idiom (synchronized methods) as in the following code:
public class SynchronizedCounter {
private int c = 0;
public synchronized void increment() {
c++;
}
public synchronized void decrement() {
c--;
}
public synchronized int value() {
return c;
}
}
You have two main effects:
1) it is not possible for two invocations of synchronized methods on the same object to interleave. 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.
2) when a synchronized method exits, it automatically establishes a happens-before relationship with any subsequent invocation of a synchronized method for the same object. This guarantees that changes to the state of the object are visible to all threads.
Another way to create synchronized code is with synchronized statements. Unlike synchronized methods, synchronized statements must specify the object that provides the intrinsic lock:
public void addName(String name) {
synchronized(this) {
lastName = name;
nameCount++;
}
nameList.add(name);
}
In your code you're using both of these idioms. Then, your first for loop doesn't need the synchronized(aObject) because your class methods are already synchronized methods.
source: http://docs.oracle.com/javase/tutorial/essential/concurrency/syncmeth.html
But let's say your class methods weren't synchronized. Your second code example:
for(A aObject : map.values()) {
A anotherReference = aObject;
synchronized(anotherReference) {
anotherReference.doSomething();
anotherReference.somethingElse();
}
}
Still works, because in Java every object has an intrinsic lock associated with it. When you call synchronized(Object o) you're acquiring the lock associated with the Object: anotherReference, that in your case is the aObject.
Let's think about two Thread: T1 and T2.
If T1 call this for loop before T2, it acquires the intrinsic lock associated with the aObject and T2 won't able to do the same till T1 will end both methods: doSomenthing() and somethinElse().
Related
I recently asked about if synchronized methods in Java shared the same lock key. The answer was yes. But the next thought I had was that this pattern is very restrictive and can cause unnecessary delays. For example suppose we have 2 synchronized methods and each of them process different and independent shared variables. What's the point of locking both methods at the same time if the processing of one variable does not affects the other? Is there a way to deal with this situation?
Suppose you have this class:
class C {
public synchronized void foo(...) { ... }
public synchronized void bar(...) { ... }
...
}
And suppose you have:
final C c_0 = new C(...);
final C c_1 = new C(...);
The fact that foo() and bar() both are synchronized does not prevent one thread from calling c_0.foo(...) while another thread simultaneously calls c_1.bar(...). In fact, it does not even prevent one thread from calling c_0.foo(...) while another thread calls c_1.foo(...).
A so-called "synchronized method"
synchronized void foo(...) { ... }
Really is just syntactic sugar for this:
void foo( ... ) {
synchronized(this) {
...
}
}
So, if one thread calls c_0.foo() it's "synchronizing" on the c_0 instance, and if another thread calls c_1.foo(), that thread is synchronizing on the c_1 instance--a different instance.
The synchronized keyword on the foo and bar methods only matters if two different threads both try to operate on the same instance at the same time.
Trying to understand Concurrency Concepts.
I saw a springboot application with a Controller Class that had 2 methods:
#RequestMapping(value = "/r1", produces = "application/json; charset=utf-8", method = RequestMethod.GET)
#ResponseBody
public ResponseEntity<> function1(...){...}
and another one in the same class as:
#RequestMapping(value = "/r2", produces = "application/json; charset=utf-8", method = RequestMethod.GET)
#ResponseBody
public synchronized ResponseEntity<>(...){....}
My question is if both methods are in the same class and since a synchronised method locks the whole object of that class, doesn't it lock the non-synchronised method as well?
Question : My question is if both methods are in the same class and since a synchronised method locks the whole object of that class, doesn't it lock the non-synchronised method as well?
Answer : No, Only the synchronized methods will be blocked while all other threads are trying to call synchronized method on same object.
Synchronized Methods From the documentation it is clear, at a time only one thread can execute the synchronized method on object, where all other thread that try executing synchronized method on same object will be blocked.
But still two threads can execute the synchronized method at a time, if they use two different objects
First, it is not possible for two invocations of synchronized methods on the same object to interleave.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.
Second, when a synchronized method exits, it automatically establishes a happens-before relationship with any subsequent invocation of a synchronized method for the same object. This guarantees that changes to the state of the object are visible to all threads.
If you have a synchronized block, you have to give it an object to lock on, like this:
Object o = new Object();
synchronized (o) {
// do something
}
If you want, you could use the object instance itself by using this instead of o, like this:
synchronized (this) {
// do something
}
If you have two methods and you want one lock that protects both methods, you could do this:
Object o = new Object();
public void methodOne() {
synchronized (o) {
// do something
}
}
public void methodTwo() {
synchronized (o) {
// do something
}
}
And like before, you can use this instead of a separate object o, like this:
public void methodOne() {
synchronized (this) {
// do something
}
}
public void methodTwo() {
synchronized (this) {
// do something
}
}
or you can achieve the same thing by marking both of those methods as sychronized, becuase synchronized methods will lock on the object instance:
public synchronized void methodOne() {
// do something
}
public synchronized void methodTwo() {
// do something
}
Non-synchronized methods will not block on anything.
Does the following mean that only ONE thread can be in ANY method of the object? Or can a multiple threads be in DIFFERENT methods just not the same one? Why?
public class SynchronizedCounter {
private int c = 0;
public synchronized void increment() {
c++;
}
public synchronized void decrement() {
c--;
}
public synchronized int value() {
return c;
}
}
Does the following mean that only ONE thread can be in ANY method of the object?
For your specific example, yes, because all the methods are synchronized and non-static.
If your class had any unsynchronized methods, then the unsynchronized methods would not be blocked.
If your class used synchronized blocks instead of methods and synchronized them on different locks, it could get much more complex.
Does the following mean that only ONE thread can be in ANY method of the object?
Yes. Non-static synchronized methods implicitly synchronize on this. It's equivalent to:
public void increment() {
synchronized(this) {
c++;
}
}
If this were a real piece of code (I know it's not), I would recommend throwing out your SynchronizedCounter class entirely and using AtomicInteger instead.
Of course, they are synchronized on the this object. If you had written your own synchronized block with different objects for the different methods you have then it won't be the case.
This is a successor of my previous question, Is this variable being safely accessed by using synchronization?
For the following program,
Class SubClassB extends SuperClassA {
protected int c;
public void inc() {
synchronized (this) {
c++;
}
}
public void dec() {
synchronized ( (SuperClassA) this) {
c--;
}
}
}
Would the counter "c" be accessed thread safe ? I am not sure that in the "dec()" method, is the SuperClassA cast "this" reference a valid object for the synchronized block ? If yes, would the two synchronized blocks lock the same "this" object ? (As it seems to me that "(SuperClassA) this" is not equal to "this")
This strange simulated code comes from the following real life example where SuperClassA is a base class that is not supposed to be modified,
Class SuperClassA {
protected int c;
public void dec() {
synchronized (this) {
c--;
}
}
}
Class SubClassB extends SuperClassA {
public void inc() {
synchronized (this) {
c++;
}
}
public void dec() {
super.dec();
}
}
In this example, the "dec()" method in SubClassB calls its superclass's "dec()" method which performs a lock to a "this" object that I suppose to be "SuperClassA.this". If the locked object in SubClassB's "inc()" method is not exactly the same as the locked object in SubClassB's "dec()" method, then I am wondering the inherited counter "c" in SubClassB may NOT be safely accessed by different threads. I feel that there is some ambiguities in using the "this" reference in synchronized blocks.
In the real life example, if I want SubClassB's counter "c" to be thread safe, do I need to add one more synchronized block in its "dec()" method, like this,
Class SubClassB extends SuperClassA {
public void inc() {
synchronized (this) {
c++;
}
}
public void dec() {
synchronized (this) {
super.dec();
}
}
}
But it seems that such added block is not elegant and may be redundant !
Does anyone have some ideas on these questions. Thanks in advance.
Lawrence
The code is thread-safe, because (SomeObject) this adn this are the same object. A cast doesn't transform an object into another object.
The code lacks encapsulation, though, because it lets any subclass access the protected c field in an unsynchronized way. So, any subclass might use c++ or c-- without any synchronization. The field should be private.
Would the counter "c" be accessed thread safe ?
Yes, it's using the same lock object.
I am not sure that in the "dec()" method, is the SuperClassA cast "this" reference a valid object for the synchronized block ?
Yes.
If yes, would the two synchronized blocks lock the same "this" object ? (As it seems to me that "(SuperClassA) this" is not equal to "this")
Yes. Even if you cast the instance to something it can be casted to (even Object), it'll still refer to the same object.
[...] But it seems that such added block is not elegant and may be redundant !
It is redundant. Extra synchronization is required only if you call multiple synchronized methods and the combined effect must be atomic.
it seems to me that "(SuperClassA) this" is not equal to "this"
Wrong; synchronization is done on objects, and casting only changes the compile-time type, is has no effect on object identity.
Thus, you don't have to add an extra synchronization in the subclass.
All three examples are correct as far as synchronization is concerned.
There's only one monitor associated with any object.
Casting this to a base class inside synchronized makes no difference.
For the same object, it doesn't matter whether synchronized(this) is invoked in the context of the derived class or the base class: the same lock is used in both cases.
What is the difference between synchronizing a static method and a non static method in java?Can anybody please explain with an example. Also is there any difference in synchronizing a method and synchronizing a block of code?
I will try and add an example to make this extra clear.
As has been mentioned, synchronized in Java is an implementation of the Monitor concept. When you mark a block of code as synchronized you use an object as a parameter. When an executing thread comes to such a block of code, it has to first wait until there is no other executing thread in a synchronized block on that same object.
Object a = new Object();
Object b = new Object();
...
synchronized(a){
doStuff();
}
...
synchronized(b){
doSomeStuff();
}
...
synchronized(a){
doOtherStuff();
}
In the above example, a thread running doOtherStuff() would block another thread from entering the block of code protecting doStuff(). However, a thread could enter the block around doSomeStuff() without a problem as that is synchronized on Object b, not Object a.
When you use the synchronized modifier on an instance method (a non-static method), it is very similar to having a synchronized block with "this" as the argument. So in the following example, methodA() and methodB() will act the same way:
public synchronized void methodA() {
doStuff();
}
...
public void methodB() {
synchronized(this) {
doStuff();
}
}
Note that if you have a methodC() in that class which is not synchronized and does not have a synchronized block, nothing will stop a thread from entering that method and careless programming could let that thread access non-safe code in the object.
If you have a static method with the synchronized modifier, it is practically the same thing as having a synchronized block with ClassName.class as the argument (if you have an object of that class, ClassName cn = new ClassName();, you can access that object with Class c = cn.getClass();)
class ClassName {
public void static synchronized staticMethodA() {
doStaticStuff();
}
public static void staticMethodB() {
synchronized(ClassName.class) {
doStaticStuff();
}
}
public void nonStaticMethodC() {
synchronized(this.getClass()) {
doStuff();
}
}
public static void unSafeStaticMethodD() {
doStaticStuff();
}
}
So in the above example, staticMethodA() and staticMethodB() act the same way. An executing thread will also be blocked from accessing the code block in nonStaticMethodC() as it is synchronizing on the same object.
However, it is important to know that nothing will stop an executing thread from accessing unSafeStaticMethodD(). Even if we say that a static method "synchronizes on the Class object", it does not mean that it synchronizes all accesses to methods in that class. It simply means that it uses the Class object to synchronize on. Non-safe access is still possible.
In short if you synchronize on a static method you will synchronize on the class (object) and not on an instance (object). That means while execution of a static method the whole class is blocked. So other static synchronized methods are also blocked.
Synchronization in Java is basically an implementation of monitors. When synchronizing a non static method, the monitor belongs to the instance. When synchronizing on a static method, the monitor belongs to the class. Synchronizing a block of code is the same idea, but the monitor belongs to the specified object. If you can get away with it, synchronized blocks are preferable because they minimize the time each thread spends in the critical section
There is virtually no difference between synchronizing a block and synchronizing a method. Basically:
void synchronized m() {...}
is the same as
void m() { synchronized(this) {...} }
By comparison a static synchronized method is the same as:
static void m() { synchronized(MyClass.class) {...} }
Dude, just a hint. Not related to your question:
If any do*Stuff() methods does either
this.a= /*yet another*/ new Object();
or
this.b= /*yet another*/ new Object();
then you are screwed. Because the lock is inside the value, not inside the reference. See Java synchronized references
From javadoc https://docs.oracle.com/javase/tutorial/essential/concurrency/locksync.html
when a static synchronized method is invoked, since a static method is associated with a class, not an object. In this case, the thread acquires the intrinsic lock for the Class object associated with the class. Thus access to class's static fields is controlled by a lock that's distinct from the lock for any instance of the class.
public static synchronized void getInstance(){}
When we acquire a lock on any class, we actually acquire a lock on "Class" class instance which is only one for all instances of class.
public synchronized void getInstance(){}
we can create multiple object's of a class and each object will have one lock associated with it.