在物体复活的情况下"The finalize method is never invoked more than once by a JVM for any given object."的含义？

Question

根据Java Doc：

Java 虚拟机对任何给定对象调用 finalize 方法的时间永远不会超过一次。

1. 上述政策可以帮助使物体不朽吗？如果我从 finalize(( 方法复活一个对象一次，那么该对象会变得不朽吗，因为 JVM 无法在该对象上第二次调用 finalize(( 方法？

或

如果对象再次符合 GC 的条件(
2. 假设指向复活对象的另一个引用符合 GC 的条件(，那么下次 JVM 将删除该对象而不为该对象调用 finalize(( 方法？

我试图通过一些示例代码来验证它，并假设第二个是正确的，但需要确认才能理解。

Answer 1

根据ernest_k 这是第二名。该对象将被垃圾回收，而无需调用其 finalize(( 方法。

添加我的示例代码以完成此答案，我创建该答案是为了验证。

import java.util.HashSet;
import java.util.Set;
class Immortal {
// making it public and non-fianl to modify from outside
public static Set<Immortal> immortals = new HashSet<>();
@Override
protected void finalize() throws Throwable {
System.out.println("Running finalize for " + this);
immortals.add(this); // Resurrect the object by creating a new reference
}
}
public class ObjectResurrection {
public static void callGC() {
// call garbage collection
System.gc();
// wait for some time to give chance to run garbage collection thread
try {
Thread.sleep(10 * 1000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public static void main(String[] args) {
Immortal immortal1 = new Immortal();
Immortal immortal2 = new Immortal();
// print objects when they are alive.
System.out.println("immortal1 = " + immortal1);
System.out.println("immortal2 = " + immortal2);
// print immortal set
System.out.println("immortal set = " + Immortal.immortals);
// make all objects garbage collectable
immortal1 = null;
immortal2 = null;
callGC();
// now objects will be null
System.out.println("immortal1 = " + immortal1);
System.out.println("immortal2 = " + immortal2);
// but stays in immortal set
System.out.println("immortal set = " + Immortal.immortals);
// remove all objects from immortal set, and make them again eligible for GC
Immortal.immortals.clear();
callGC();
// again print the immortal set
// this time set will be empty, and those 2 objects will be destroyed by Garbage
// Collection,
// but this time they will not call finalize, because it is already called
// you can notice no output from finalize method in the output this time
System.out.println("immortal set = " + Immortal.immortals);
}
}

上述程序的输出(在我的机器上(：

immortal1 = effective.java.item7.finalizer6.Immortal@70dea4e
immortal2 = effective.java.item7.finalizer6.Immortal@5c647e05
immortal set = []
Running finalize for effective.java.item7.finalizer6.Immortal@5c647e05
Running finalize for effective.java.item7.finalizer6.Immortal@70dea4e
immortal1 = null
immortal2 = null
immortal set = [effective.java.item7.finalizer6.Immortal@5c647e05, effective.java.item7.finalizer6.Immortal@70dea4e]
immortal set = []