关于Java敲定方法的面试题

Question

I have came across difficult (for me) Java interview question on finalize method. Suppose you have given finalize method as shown below:

public void finalize()
{
     a.b = this;
}

Now the following object scenario is given.

How would you solve this problem? If A was not referring to B then this problem could be easier as GC will run, it will collect B and call finalize for B but here A is referring B so its difficult. How finalize will work in this scenario?

Any ideas? Thanks in advance

Answer 1

Very interesting question. From the JDK1.6 Doc, I find these two sentences: 1. The finalize method may take any action, including making this object available again to other threads 2. The finalize method is never invoked more than once by a Java virtual machine for any given object.

So in my opinion, for the first time, when B is collecting by GC, finalize method will be invoked if A is still available to some threads then B becomes available again, this time the GC will not collect B. But because finalize method will be invoked only once, so next time when the GC find B can't be accessed by any thread then GC will collect B.

Answer 2

The easiest way to think of Java finalization is to consider it an extra 'bit' of state that every object with a finalizer has. When a new object is created, this isFinalized bit is set to false. When the garbage collector finds that an object with a finalizer is unreachable, it checks this isFinalized bit and only reclaims the object if it is true -- if it is false it instead runs the finalizer and sets the bit to true. Once set, there's no way for the bit to ever be cleared, so in any later time the garbage collector runs, if it's unreachable, it will be collected.

Answer 3

It is not entirely clear what you are asking.

However, the JLS 12.6.2 states that the order in which objects are finalized is not specified. This means that the finalize methods for all finalizable classes should be designed to work when invoked in any order.

Note that this applies equally to unreachable and finalizer-reachable objects. In other words, the 3 objects in your diagram could be finalized in any order.

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... how B will be garbage collected as A is reachable and how finalize will get called for B?

Perhaps it is a trick question. If A is reachable, then B is also reachable, so it won't be garbage collected, and its finalize method won't be called. If A becomes unreachable then so does B, and both will be finalized.

What that finalize method will actually do depends on what class it belongs to. Lets assume it is a method of the "guaranteed reachable object", and the a variable contains a reference to A:

• The finalize method won't be called by the GC because the object is reachable.

• Some other code could explicitly call that finalize method. If that happened, then ab will no longer refer to B, and B will be unreachable and eligible for eventual garbage collection, finalization and (ultimately) deletion.

Answer 4

Sorry if I am wrong. But to me, B is a field of A, so A definitely need to keep a reference to B, and while A "always-live" object holding a reference to A, both A and B will not be GC and the problem to GC B is just not making any sense.