blocks|key|561816|text|来自Object.hashCode文档：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|561817|hashCode的总合同是：|561818|每当在执行Java应用程序期间在同一个对象上多次调用它时，hashCode方法必须始终如一地返回相同的整数，前提是不修改关于该对象的相等比较中使用的任何信息。从应用程序的一次执行到同一应用程序的另一次执行，这个整数不需要保持一致。|unordered-list-item|561819|如果两个对象根据equals(Object)方法相等，那么对两个对象中的每个对象调用hashCode方法都必须产生相同的整数结果。|561820|如果两个对象根据等于(java.lang.Object)方法是不相等的，则不需要对每个对象调用hashCode方法就必须产生不同的整数结果。但是，程序员应该意识到，为不平等的对象生成不同的整数结果可能会提高哈希表的性能。|561821|因此，对象始终具有相同的hashCode是有效的。|561822|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Object.html#hashCode%2528%2529^0|2|F|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$]]|$1|I|3|J|5|H|7|12|8|@]|9|@]|C|$]]|$1|K|3|L|5|H|7|13|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|14|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

From the <a href="http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Object.html#hashCode%28%29" rel="nofollow">Object.hashCode</a> documentation:

The general contract of hashCode is:

<ul>
<li>Whenever it is invoked on the same object more than once during an
execution 
 of a Java application, the hashCode method must consistently return the same integer, provided no information used in equals
comparisons on the object is modified. This integer need not remain
consistent from one execution of an application to another execution
of the same application.</li>
<li>If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must
produce the same integer result.</li>
<li>It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method
on each of the two objects must produce distinct integer results. However, the programmer should be aware that producing distinct integer results for unequal objects may improve the performance of hashtables. </li>
</ul>

Therefore it is valid that your Object has the same hashCode all the time.

blocks|key|3045418|text|如果System.out.println找到对象引用，它总是打印toString()方法。这里，List用不同的toString实现存储不同的对象引用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3045419|+ArrayList<Object>+a+=+new+ArrayList<Object>();
++++a.add(new+Socket());|code-block|syntax|javascript|3045420|例如，Socket类toString被重写为|3045421|public+String+toString()+{
++++try+{
++++++++if+(isConnected())
++++++++++++return+"Socket[addr="+%2B+getImpl().getInetAddress()+%2B
++++++++++++++++",port="+%2B+getImpl().getPort()+%2B
++++++++++++++++",localport="+%2B+getImpl().getLocalPort()+%2B+"]";
++++}+catch+(SocketException+e)+{
++++}
++++return+"Socket[unconnected]";
++}|3045422|因此，您将看到new+Socket()的输出为Socket[unconnected]。|3045423|entityMap^0|2|I|W|A|1C|4|1K|8|0|0|3|6|A|8|0|0|7|C|N|J|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|R|$]]

If <code>System.out.println</code> find Object reference than it always print <code>toString()</code> method. Here the <code>List</code> store differents Object reference with different <code>toString</code> implementations.

<pre><code> ArrayList&lt;Object&gt; a = new ArrayList&lt;Object&gt;();
 a.add(new Socket());
</code></pre>

For example, The <code>Socket</code> class <code>toString</code> is overrided as

<pre><code>public String toString() {
 try {
 if (isConnected())
 return "Socket[addr=" + getImpl().getInetAddress() +
 ",port=" + getImpl().getPort() +
 ",localport=" + getImpl().getLocalPort() + "]";
 } catch (SocketException e) {
 }
 return "Socket[unconnected]";
 }
</code></pre>

So, you will see the output of <code>new Socket()</code> as <code>Socket[unconnected]</code>

blocks|key|3045452|text|对象的哈希代码通常是从其内部地址(ref+http://docs.oracle.com/javase/7/docs/api/java/lang/Object.html#hashCode%2528%2529)计算的，因此，如果您以相同的状态重新运行程序，则内存中对象的地址可能每次都是相同的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3045453|从技术上讲，它并不是重用同一个对象，因为您正在退出JVM；它每次都遵循相同的步骤，在内存中相同的位置创建一个相同的对象，这将产生一个相同的哈希代码供toString()返回。|3045454|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/lang/Object.html#hashCode%2528%2529^0|L|25|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

An object's hash code is typically calculated from its internal address (ref <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/Object.html#hashCode%28%29">http://docs.oracle.com/javase/7/docs/api/java/lang/Object.html#hashCode%28%29</a>) so if you're re-running the program with an identical state, the address of the object in memory is likely to be the same each time.

It's not technically reusing the same object, because you're exiting the JVM; it's just following the same steps each time, creating an identical object in the same place in memory, which produces an identical hash code for toString() to return.

blocks|key|1965512|text|与以前一样，新对象恰好是在同一个内存位置创建的，很少执行。预计情况会有所改变。您可能会尝试执行程序10次，我觉得至少2-3次对象将被创建在不同的位置。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1965513|不能声称这种JVM行为是错误的。甚至查看java中的内存位置也是错误的，JVM可以在执行过程中将对象移动到不同的内存位置。|1965514|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

New Object just happens to be created in the same memory location as before with few executions. It is expected to change. You might try executing the program 10 times, I feel atleast 2-3 times the object will be created in a different location.

This JVM behaviour cannot be claimed as wrong. It is wrong to even look into memory location in java and JVM has options to move the objects to different memory locations during the course of the execution.

blocks|key|1965548|text|是的，就像你说的那样，我试过几次执行Java应用程序。结果与您的said.but相同，在我们执行application.however时得到的结果与在eclipse中重新编译应用程序时的结果不一样。像这样，您将得到不同的哈希码values.Firstly，当第一个执行+java+program.every时间执行将得到相同的哈希代码值时，JVM将将字节文件加载到内存中，就像答复的"Rich“一样。每次编译程序时，都会将字节文件加载到内存中。因此，我们将得到不同的哈希码值和不同的对象。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1965549|entityMap^0|22|D|3O|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

yeah,just like do this what you said.i tried to execute the Java application several times. the result the same as you said.but,the results just when we executed Java application.however,it's not same results when we recompile the Java application in eclipse. like this,you will get different hashcode values.Firstly,JVM will load the bytes file to memory when first execute java program.every time execute will get the same hashcode value ,just like "Rich Smith" replyed. every times will load the bytes file to memory when compiled program . so we will get different hashcode value and get different Object.

I am having one of those weird moments. 

<pre><code>ArrayList&lt;Object&gt; a = new ArrayList&lt;Object&gt;();
a.add(new Socket());
a.add(new Thread());
a.add("three");
a.add(a);
a.add(new Object());

for(Object output : a) {
 System.out.println(output);
}
</code></pre>

Output:

<pre><code>Socket[unconnected]
Thread[Thread-0,5,main]
three
[Socket[unconnected], Thread[Thread-0,5,main], three, (this Collection)]
java.lang.Object@615e7597
</code></pre>

Every time I run this, the new object always gives the same Hex String (Java doc <code>Integer.toHexString(hashCode())</code>), why is this? Why doesn't this produce a different String each time? Or is it reusing the same object because it can?

EDIT: I tried executing the Java application several times.

`new Object()' does not seem to create a new object, why?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在经历这样一个奇怪的时刻。ArrayList<Object> a = new ArrayList<Object>();a.add(new Socket());a.add(new Thread());a.add("three");a.add(a);a.add(new Object());for(Object outp...

问‘()’似乎没有创建一个新对象，为什么？
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问‘()’似乎没有创建一个新对象，为什么？EN