Immutable objects are thread safe, but why?

问题:

If one thread is creating populating the reference variable of the immutable class by creating its object and at the second time the other thread kicks in before the first thread completes and creates another object of the immutable class, won't the immutable class usage be thread unsafe? Creating a immutable object also says that all fields to marked as final as..... "it may be necessary to ensure correct behavior if a reference to a newly created instance is passed from one thread to another without synchronization" Are they trying to say that the other thread may re-point the reference variable to some other object of the immutable class and that way the threads will be pointing to different objects leaving the state inconsistent?

回答1:

Immutable objects are thread safe, but why?

An immutable object is an object that is no longer modified once it has been constructed. If in addition, the immutable object is only made accessible to other thread after it has been constructed, and this is done using proper synchronization, all threads will see the same valid state of the object.

If one thread is creating populating the reference variable of the immutable class by creating its object and at the second time the other thread kicks in before the first thread completes and creates another object of the immutable class, won't the immutable class usage be thread unsafe?

No. What makes you think so? An object's thread safety is completely unaffected by what you do to other objects of the same class.

Are they trying to say that the other thread may re-point the reference variable to some other object of the immutable class and that way the threads will be pointing to different objects leaving the state inconsistent?

They are trying to say that whenever you pass something from one thread to another, even if it is just a reference to an immutable object, you need to synchronize the threads. (For instance, if you pass the reference from one thread to another by storing it in an object or a static field, that object or field is accessed by several threads, and must be thread-safe)

回答2:

Thread safety is data sharing safety, And because in your code you make decisions based on the data your objects hold, the integrity and deterministic behaviour of it is vital. i.e

Imagine we have a shared boolean instance variable across two threads that are about to execute a method with the following logic

• If flag is false, then I print "false" and then I set the flag back to true.
• If flag is true, then I print "true" and then I set the flag back to false.

If you run continuously in a single thread loop, you will have a deterministic output which will look like:

false - true - false - true - false - true - false ...

But, if you ran the same code with two threads, then, the output of your output is not deterministic anymore, the reason is that the thread A can wake up, read the flag, see that is false, but before it can do anything, thread B wakes up and reads the flag, which is also false!! So both will print false... And this is only one problematic scenario I can think of... As you can see, this is bad.

If you take out the updates of the equation the problem is gone, just because you are eliminating all the risks associated with data sync. that's why we say that immutable objects are thread safe.

It is important to note though, that immutable objects are not always the solution, you may have a case of data that you need to share among different threads, in this cases there are many techniques that go beyond the plain synchronization and that can make a whole lot of difference in the performance of your application, but this is a complete different subject.

Immutable objects are important to guarantee that the areas of the application that we are sure that don't need to be updated, are not updated, so we know for sure that we are not going to have multithreading issues

You probably might be interested in taking a look at a couple of books:

This is the most popular: http://www.amazon.co.uk/Java-Concurrency-Practice-Brian-Goetz/dp/0321349601/ref=sr_1_1?ie=UTF8&qid=1329352696&sr=8-1

But I personally prefer this one: http://www.amazon.co.uk/Concurrency-State-Models-Java-Programs/dp/0470093552/ref=sr_1_3?ie=UTF8&qid=1329352696&sr=8-3

Be aware that multithreading is probably the trickiest aspect of any application!

回答3:

In addition to other answers posted already, immutable objects once created, they cannot be modified further. Hence they are essentially read-only.

And as we all know, read-only things are always thread-safe. Even in databases, multiple queries can read same rows simultaneously, but if you want to modify something, you need exclusive lock for that.

回答4:

Actually immutable objects are always thread-safe, but its references may not be.

Confused?? you shouldn't be:-

Going back to basic: Thread-safe simply means that two or more threads must work in coordination on the shared resource or object. They shouldn't over-ride the changes done by any other thread.

Now String is an immutable class, whenever a thread tries to change it, it simply end up creating a new object. So simply even the same thread can't make any changes to the original object & talking about the other thread would be like going to Sun but the catch here is that generally we use the same old reference to point that newly created object.

When we do code, we evaluate any change in object with the reference only.

Statement 1: String str = "123"; // initially string shared to two threads

Statement 2: str = str+"FirstThread"; // to be executed by thread one

Statement 3: str=str+"SecondThread"; // to be executed by thread two

Now since there is no synchronize, volatile or final keywords to tell compiler to skip using its intelligence for optimization (any reordering or caching things), this code can be run in following manner.

1. Load Statement2, so str = "123"+"FirstThread"
2. Load Statement3, so str = "123"+"SecondThread"
3. Store Statement3, so str = "123SecondThread"
4. Store Statement2, so str = "123FirstThread"

and finally the value in reference str="123FirstThread" and for sometime if we assume that luckily our GC thread is sleeping, that our immutable objects still exist untouched in our string pool.

So, Immutable objects are always thread-safe, but their references may not be. To make their references thread-safe, we may need to access them from synchronized blocks/methods.

回答5:

Two threads will not be creating the same object, so no problem there.

With regards to 'it may be necessary to ensure...', what they are saying is that if you DON'T make all fields final, you will have to ensure correct behavior yourself.

回答6:

Immutability doesn't imply thread safety.In the sense, the reference to an immutable object can be altered, even after it is created.

//No setters provided class ImmutableValue {       private final int value = 0;       public ImmutableValue(int value)      {           this.value = value;      }       public int getValue()      {           return value;      } }  public class ImmutableValueUser{   private ImmutableValue currentValue = null;//currentValue reference can be changed even after the referred underlying ImmutableValue object has been constructed.    public ImmutableValue getValue(){     return currentValue;   }    public void setValue(ImmutableValue newValue){     this.currentValue = newValue;   }  } 

回答7:

Immutability doesn't ensure thread safety ,if threads are just accessing immutable objects to read then you don't have to take extra precaution to make sure that the data is consistent,but if threads can manipulate data then you have to use some safe way to share data among threads so that each thread sees consistent data.

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