前言:承接上一篇hashCode和identityHashCode 的关系,下面的两段简单的程序主要是检验一下系统生成的identityHashCode是否存在重复的情况。
1:可以自由控制生成对象的个数,并且不受测试的类是否重写hashCode()方法的影响
import java.util.HashSet; import java.util.Set; public class CheckSystemIdentity { public static void main(String args[]) { //声明set对象 Set<Integer> hashSet = new HashSet<Integer>(1024); //通过循环遍历,检查生成的hashCode是否存在重复的现象 int colls = 0; int cycleSize=1000000; for (int n = 0; n < cycleSize; n++) { Integer obj = new Integer(666); int identityHashCode = System.identityHashCode(obj); //System.out.println("identityHashCode is : "+identityHashCode); Integer hashValue = Integer.valueOf(identityHashCode); //System.out.println("hashValue is : "+hashValue+"\n"); if (hashSet.contains(hashValue)) { System.err.println("System.identityHashCode() collision!"); colls++; } else { hashSet.add(hashValue); } } //System.out.println("Integer.MAX_VALUE is : "+Integer.MAX_VALUE+"\n"); System.out.println("created "+cycleSize+" different objects - " + colls + " times with the same value for System.identityHashCode()"); } }
2:利用死循环来检测系统生成的identityHashCode是否存在重复的情况
2-1:测试类是自定义的,没有重写hashCode()方法
import java.util.Hashtable; import java.util.Map; public class HashCodeTest{ //试验对象,没有重写hashCode()方法 static class DummyObject{ } public static void reportClash(DummyObject obj1, DummyObject obj2) { System.out.println("obj1.hashCode() = " + obj1.hashCode()); System.out.println("obj2.hashCode() = " + obj2.hashCode()); System.out.println("(obj1 == obj2) = " + (obj1 == obj2) + " (!)"); } public static void main(String[] args) { Map<Integer,DummyObject> map = new Hashtable<Integer ,DummyObject>(); //通过死循环,检查生成的hashCode是否存在重复的情况 for (int count = 1; true; count++) { DummyObject obj = new DummyObject(); if (map.containsKey(obj.hashCode())) { System.out.println("Clash found after instantiating " + count + " objects."); reportClash(map.get(obj.hashCode()), obj); System.exit(0); } System.out.println("The method execute " + count + " and the object hashCode is "+obj.hashCode()); map.put(obj.hashCode(), obj); } } }
2-2:测试类是String,重写了hashCode()方法和2-1正好再次的做一下对比
import java.util.Hashtable; import java.util.Map; public class HashCodeTest { public static void reportClash(String obj1, String obj2) { System.out.println("obj1.hashCode() = " + obj1.hashCode()); System.out.println("obj2.hashCode() = " + obj2.hashCode()); System.out.println("(obj1 == obj2) = " + (obj1 == obj2) + " (!)"); } public static void main(String[] args) { Map<Integer,String> map = new Hashtable<Integer ,String>(); //通过死循环,检查生成的hashCode是否存在重复的情况 for (int count = 1; true; count++) { String obj = new String(); if (map.containsKey(obj.hashCode())) { System.out.println("Clash found after instantiating " + count + " objects."); reportClash(map.get(obj.hashCode()), obj); System.exit(0); } System.out.println("The method execute " + count + " and the object hashCode is "+obj.hashCode()); map.put(obj.hashCode(), obj); } } }
3:程序相对简单,有兴趣的可以自己运行一下看看结果。
结论如下:
3-1:在我实验的环境中(win7+jdk7)identityHashCode没有出现重复的现象
3-2:没有重写的hashCode和identityHashCode是一致的,可以间接的反映一个对象的内存地址是什么
3-3:identityHashCode能更为准确的代表一个对象和其内存地址的hash关系,
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