重写equals()方法就必须重写hashCode()方法的原因

一.Object类中关于hashCode()方法的注释：

/**
* Returns a hash code value for the object. This method is
* supported for the benefit of hashtables such as those provided by
* <code>java.util.Hashtable</code>.
* <p>
* The general contract of <code>hashCode</code> is:
* <ul>
* <li>Whenever it is invoked on the same object more than once during
*     an execution of a Java application, the <tt>hashCode</tt> method
*     must consistently return the same integer, provided no information
*     used in <tt>equals</tt> 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>If two objects are equal according to the <tt>equals(Object)</tt>
*     method, then calling the <code>hashCode</code> method on each of
*     the two objects must produce the same integer result.
* <li>It is <em>not</em> required that if two objects are unequal
*     according to the {@link java.lang.Object#equals(java.lang.Object)}
*     method, then calling the <tt>hashCode</tt> 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.
* </ul>
* <p>
* As much as is reasonably practical, the hashCode method defined by
* class <tt>Object</tt> does return distinct integers for distinct
* objects. (This is typically implemented by converting the internal
* address of the object into an integer, but this implementation
* technique is not required by the
* Java<font size="-2"><sup>TM</sup></font> programming language.)
*
* @return  a hash code value for this object.
* @see     java.lang.Object#equals(java.lang.Object)
* @see     java.util.Hashtable
*/
public native int hashCode();

简单翻译一下，大概意思就是

1.在 Java 应用程序执行期间，在对同一对象多次调用 hashCode 方法时，必须一致地返回相同的整数，前提是将对象进行 equals 比较时所用的信息没有被修改。从某一应用程序的一次执行到同一应用程序的另一次执行，该整数无需保持一致。

2.如果根据 equals(Object) 方法，两个对象是相等的，那么对这两个对象中的每个对象调用 hashCode 方法都必须生成相同的整数结果。

3.如果根据 equals(java.lang.Object) 方法，两个对象不相等，那么对这两个对象中的任一对象上调用 hashCode 方法不 要求一定生成不同的整数结果。但是，程序员应该意识到，为不相等的对象生成不同整数结果可以提高哈希表的性能。

二、重写的意义

hashcode是用于散列数据的快速存取，如利用HashSet/HashMap/Hashtable类来存储数据时，都是根据存储对象的hashcode值来进行判断是否相同的。

这样如果我们对一个对象重写了euqals，意思是只要对象的成员变量值都相等那么euqals就等于true，但不重写hashcode，那么我们再new一个新的对象，

当原对象.equals（新对象）等于true时，两者的hashcode却是不一样的，由此将产生了理解的不一致，如在存储散列集合时（如Set类），将会存储了两个值一样的对象，

导致混淆，因此，就也需要重写hashcode()

三、重写hashCode()的原则

1.同一个对象多次调用hashCode()方法应该返回相同的值；

2.当两个对象通过equals()方法比较返回true时，这两个对象的hashCode()应该返回相等的（int）值；

3.对象中用作equals()方法比较标准的Filed(成员变量（类属性）)，都应该用来计算hashCode值。

查看String源码，看hashCode()d的实现方法：

/**
* Returns a hash code for this string. The hash code for a
* <code>String</code> object is computed as
* <blockquote><pre>
* s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
* </pre></blockquote>
* using <code>int</code> arithmetic, where <code>s[i]</code> is the
* <i>i</i>th character of the string, <code>n</code> is the length of
* the string, and <code>^</code> indicates exponentiation.
* (The hash value of the empty string is zero.)
*
* @return  a hash code value for this object.
*/
public int hashCode() {
int h = hash;
if (h == 0 && value.length > 0) {
char val[] = value;

for (int i = 0; i < value.length; i++) {
h = 31 * h + val[i];
}
hash = h;
}
return h;
}