blocks|key|705502|text|根据定义，静态变量不是实例变量，因此在同一个类的所有实例中总是相等的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|705503|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Static variables, by definition, are not instance variables, and will therefore always be equal across all instances of the same class.

blocks|key|4986002|text|您将比较每个实例变量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4986003|静态变量保证是相等的，因为它们是特定于类的，而不是特定于实例的，所以您不必担心比较它们。|4986004|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|$]]

You would compare every instance variable. 

Static variables are guaranteed to be equal, as they are class-specific, not instance-specific, so you don't have to worry about comparing them.

blocks|key|706111|text|绝对不是静态的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|706112|是否比较所有实例变量取决于是什么决定了对象的“身份”，也就是说，什么时候你认为它们是相等的？这只能在您的特定应用程序的上下文中决定-我们需要更多信息。|706113|例如，如果您有一个表示书籍的类，如果您只想存储关于两本书的元数据(标题、作者)，则可能只比较ISBN编号来确定两本书是否为同一本书。如果您合并了两个这样的数据库，您可能希望消除重复记录。|706114|但是，如果您正在实现实际实体图书的图书馆目录，则每个副本都很重要且各不相同，因此您可以比较ISBN和副本数量。如果您合并了两个库，您应该能够检测到重复的副本。|706115|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

Definitely not the static ones. 

Whether you compare all the instance variables depends on what determines the "identity" of your objects, i.e. when do you consider them equal? This can only be decided in the context of your particular application - we'd need more information.

For example, if you had a class representing books, you might only compare the ISBN number to determine whether two books are the same book, if you just wanted to store metadata about them (title, author). If you merged two such databases, you'd want to eliminate duplicate records.

BUT, if you were implementing a library catalogue of actual physical books, each individual copy is important and different, so you might compare ISBN and copy number. If you merged two libraries, you'd expect to be able to detect duplicate copies.

blocks|key|948678|text|这是没有意义的，因为使用equals()可以比较实例，还因为它们是静态的:您会将它们与自身进行比较吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|948679|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It would make no sense, both because with equals() you compare instances and because, well, they are static: what would you compare them to, themselves?

blocks|key|1721323|text|比较静态数据是没有用的，因为静态数据是在Die类的所有实例之间共享的。但是您可以通过直接访问不同的字段来比较它们(参见下面的示例)。请注意，如果您的Die对象具有复杂的字段(例如Map、Set等的实例)，则还应该对这些对象调用equals方法(同样，请参见下面的示例)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1721324|如果希望提供equals+()方法，则应该覆盖Object类(equals+(Object+anOtherObject)中提供的方法，而不是重载它，或者至少确保还覆盖了equals+(Object+anOtherObject)，以确保它也返回正确的值(默认实现只检查它是否是同一个实例)。在你的方法中，你应该检查anOtherObject是否是Die的一个实例。|1721325|下面是一个示例，假设您的Die类有3个字段：String+name、int+value和Map<Integer>+complexField|1721326|public+boolean+equals+(Object+anOtherObject)+{
++++if+(anOtherObject+==+this)+{
++++++++return+true;
++++}

++++if+(!anOtherObject+instanceof+Die)+{
++++++++return+false;
++++}

++++Die+otherDie+=+(Die)+anOtherObject;

++++if+(this.value+!=+otherDie.value+%7C%7C
++++++++++++!this.name.equals+(otherDie.name)+%7C%7C
++++++++++++!this.complexField.equals+(otherDie.complexField))+{
++++++++return+false;
++++}

++++return+true;
}|code-block|syntax|javascript|1721327|Josh+Bloch的"Effective“中有一个关于如何正确实现equals的非常详细的部分。你一定要读一读。|1721328|entityMap^0|K|3|22|3|2H|3|2L|3|35|6|0|6|9|N|6|V|S|2C|T|4C|D|4S|3|0|C|3|M|B|Y|9|18|P|0|0|Y|6|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]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|1P|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1Q|8|@$9|1R|A|1S|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]]|R|$]]

It is useless to compare static data, as it is shared among all instances of your <code>Die</code> class. But you can compare the various fields by accessing them directly (see example below). Note that if your <code>Die</code> object has complex fields (such as instances of <code>Map</code>, <code>Set</code>, etc), you should also call the <code>equals</code> methods on those objects (again, see example below).

If you want to provide an <code>equals ()</code> method, you should override the one provided in the <code>Object</code> class (<code>equals (Object anOtherObject</code>), rather than overloading it, or at least make sure you also override <code>equals (Object anOtherObject)</code> to make sure that it returns the correct value too (the default implementation only checks if it's the same instance). And in your method you should then check if <code>anOtherObject</code> is an instance of <code>Die</code>. 

Here's an example, assuming that your <code>Die</code> class has 3 fields: <code>String name</code>, <code>int value</code> and <code>Map&lt;Integer&gt; complexField</code>:

<pre><code>public boolean equals (Object anOtherObject) {
 if (anOtherObject == this) {
 return true;
 }

 if (!anOtherObject instanceof Die) {
 return false;
 }

 Die otherDie = (Die) anOtherObject;

 if (this.value != otherDie.value ||
 !this.name.equals (otherDie.name) ||
 !this.complexField.equals (otherDie.complexField)) {
 return false;
 }

 return true;
}
</code></pre>

Josh Bloch's "Effective Java" has a very details section on how to correctly implement <code>equals</code>. You should definitely read it.

blocks|key|1721347|text|下模布尔值等于(+|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1721348|+aDie)|blockquote|1721349|1721350|1721351|是错误的，则类将调用equals(Object)方法并忽略您的等于(Die)。还要使用equals(Object)使用的相同字段实现int+hashCode()方法。|offset|length|style|CODE|1721352|+@Override+public+boolean+equals(Object+aDie){
+++++if(aDie+==+null+%7C%7C+aDie.getClass()+!=+Die.class)return+false;
+++++if(aDie+==+this)return+true;
+++++Die+other+=+(Die)aDie;
+++++...+++
+}
+@Override+public+int+hashCode(){
+++++...
+}|code-block|syntax|javascript|1721353|您可以忽略静态字段，因为它们对于每个骰子都是相同的。|1721354|entityMap^0|0|0|0|0|A|E|1U|E|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$]]|$1|E|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Z|8|@$I|10|J|11|K|L]|$I|12|J|13|K|L]]|9|@]|A|$]]|$1|M|3|N|5|O|7|14|8|@]|9|@]|A|$P|Q]]|$1|R|3|S|5|6|7|15|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|U|$]]

<blockquote>
 boolean equals(Die aDie)
</blockquote>

is wrong, classes will call the <code>equals(Object)</code> method and ignore your equals(Die). Also implement the <code>int hashCode()</code> method using the same fields that equals(Object) uses.

<pre><code> @Override public boolean equals(Object aDie){
 if(aDie == null || aDie.getClass() != Die.class)return false;
 if(aDie == this)return true;
 Die other = (Die)aDie;
 ... 
 }
 @Override public int hashCode(){
 ...
 }
</code></pre>

You can ignore static fields since they are the same for every Die.

blocks|key|706203|text|当一个变量用关键字“static”声明时，它被称为“类变量”。所有实例共享变量的相同副本(在同一类的所有实例中始终相等)。可以通过类直接访问类变量，而不需要创建实例。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|706204|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

When a variable is declared with the keyword “static”, its called a “class variable”. All instances share the same copy of the variable (always be equal across all instances of the same class). A class variable can be accessed directly with the class, without the need to create a instance.

I am creating the method .equals(Die aDie) in my program. Do I compare every instance variable including static ones?

Implementing .equals(Die aDie) method and static variables

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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我正在我的程序中创建方法.equals(Die aDie)。我是否要比较每个实例变量，包括静态变量？

问实现.equals(Die aDie)方法和静态变量
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