如果我比较两个A的实例而不覆盖equals方法，我会得到预期的结果吗？ / p>

这取决于你的期望：）

默认实现将给出你引用相等 - 换句话说，当你比较两个引用时， equals 只有在它们引用同一个对象时才会返回true。 / p>

您通常会覆盖等于来实现值相等 - 其中两个不同的对象被认为是相等的，通常是通过美德具有相等的字段值本身。相等的确切含义取决于您的设计 - 例如，两个对象仍然可以通过其他方式区分。

如果覆盖等于，你应该也覆盖 hashCode 以与等于一致>，如果 a.equals（b）为真，那么 a.hashCode（）== b.hashCode（）。这将允许您的类的实例用作基于散列的集合中的键（例如 HashMap ），以便您可以根据等于原始的，而不是必须使用对完全原始密钥对象的引用。

Possible Duplicate:
Overriding equals and hashCode in Java

If I have

class A {
    int x = 1;
}
...
A a1 = new A();
A a2 = new A();
a1.equals(a2);

If I compare 2 instances of A without override the equals method, will I get expected result?

If I compare 2 instances of A without override the equals method, will I get expected result?

That depends on what you expect :)

The default implementation will give you reference equality - in other words, when you compare two references, equals will only return true if they're references to the same object.

You would normally override equals to implement "value equality" - where two distinct objects are deemed equal, usually by virtue of having equal field values themselves. The exact meaning of equality will depend on your design - the two objects could still be distinguishable in other ways, for example.

If you override equals, you should also override hashCode to be consistent with equals, such that if a.equals(b) is true, then a.hashCode() == b.hashCode(). This will allow instances of your class to be used as keys in hash-based collections (e.g. HashMap) so that you can look up a value based on a key which is equal to the original one, rather than having to use a reference to the exact original key object.