equals是Object的公有方法，那么我们通常都会在自己的类中重写这个equals方法，同时必须重写hasCode方法，知道为什么重写equals方法必须重写hasCode方法呢？

/** * Returns a hash code value for the object. This method is * supported for the benefit of hash tables such as those provided by * {@link java.util.HashMap}. * <p> * The general contract of {@code hashCode} is: * <ul> * <li>Whenever it is invoked on the same object more than once during * an execution of a Java application, the {@code hashCode} method * must consistently return the same integer, provided no information * used in {@code 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>If two objects are equal according to the {@code equals(Object)} * method, then calling the {@code hashCode} 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 {@code 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 hash tables. * </ul> * <p> * As much as is reasonably practical, the hashCode method defined by * class {@code Object} 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™ programming language.) * * @return a hash code value for this object. * @see java.lang.Object#equals(java.lang.Object) * @see java.lang.System#identityHashCode */ public native int hashCode(); /** * Indicates whether some other object is "equal to" this one. * <p> * The {@code equals} method implements an equivalence relation * on non-null object references: * <ul> * <li>It is <i>reflexive</i>: for any non-null reference value * {@code x}, {@code x.equals(x)} should return * {@code true}. * <li>It is <i>symmetric</i>: for any non-null reference values * {@code x} and {@code y}, {@code x.equals(y)} * should return {@code true} if and only if * {@code y.equals(x)} returns {@code true}. * <li>It is <i>transitive</i>: for any non-null reference values * {@code x}, {@code y}, and {@code z}, if * {@code x.equals(y)} returns {@code true} and * {@code y.equals(z)} returns {@code true}, then * {@code x.equals(z)} should return {@code true}. * <li>It is <i>consistent</i>: for any non-null reference values * {@code x} and {@code y}, multiple invocations of * {@code x.equals(y)} consistently return {@code true} * or consistently return {@code false}, provided no * information used in {@code equals} comparisons on the * objects is modified. * <li>For any non-null reference value {@code x}, * {@code x.equals(null)} should return {@code false}. * </ul> * <p> * The {@code equals} method for class {@code Object} implements * the most discriminating possible equivalence relation on objects; * that is, for any non-null reference values {@code x} and * {@code y}, this method returns {@code true} if and only * if {@code x} and {@code y} refer to the same object * ({@code x == y} has the value {@code true}). * <p> * Note that it is generally necessary to override the {@code hashCode} * method whenever this method is overridden, so as to maintain the * general contract for the {@code hashCode} method, which states * that equal objects must have equal hash codes. * * @param obj the reference object with which to compare. * @return {@code true} if this object is the same as the obj * argument; {@code false} otherwise. * @see #hashCode() * @see java.util.HashMap */ public boolean equals(Object obj) { return (this == obj); }

上面是Object对象中hasCode和equals的签名和方法说明。

判断一些对象是否等于当前对象，实现在非空对象映射的等价关系。this==obj,由此可知，这是一个引用对象的等价判断，只有在this和obj是指向用一个应用时才返回true.这是判定实例的唯一性，我们通常使用equals时，其实并不想判断实例的唯一性，而是想要比较里面的几个（自定义）属性值是否相等，来判断是否为同一个"东西"。

契约：
reflexive(自反性)：x为非null时，x.equals(x),一定返回true.
symmetric(对称性):x,y都不为null，如果x.equals(y)返回true,那么y.equals(x).
transitive(传递性):x,y,z都不为null,如果x.equals(y),y.equals(z)返回true,那么x.equals(z)返回true.
consistent(一致性):对于任何非null的x\y，在没有修改这些对象信息的前提下，多次调用下，x.equals(y)结果是一致的。
非null的x,x.equals(null)返回false.

下面是Book类的equals重写实现，通过bookN、bookName确定是否为同一本书（东西），而不是判断两个对象的引用是否为同一个。

@Override public boolean equals(Object o) { if (this == o) return true; if (o == null || getClass() != o.getClass()) return false; Book book = (Book) o; return Objects.equal(bookNo, book.bookNo) && Objects.equal(bookName, book.bookName); }为啥重写hasCode方法

hasCode锲约：
1.任何时候在没有修改对象的前提下，多次调用同一个对象的hasCode方法，这个返回值必须是同一个Integer值。
2.如果两个对象equals结果为true，那么hasCode的值必须相同。
3.如果两个对象equals结果为false,那么hasCode就不一定不相同。

如果重写equals方法时，不去重写hasCode方法，那么必然会违背第二条锲约。所以每次hasCode必须跟随equals重写而重写。

是不是有人就是问：如果就不重写hasCode，不行吗？就违背锲约不可以吗？只equals重写有什么问题吗？
那么这里就会引申出hasCode在集合中的应用场景。

Java中的集合（Collection）有两类，一类是List，再有一类是Set。前者集合内的元素是有序的，元素可以重复；后者元素无序，但元素不可重复。那么这里就有一个比较严重的问题了：要想保证元素不重复，可两个元素是否重复应该依据什么来判断呢？这就是 Object.equals方法了。但是，如果每增加一个元素就检查一次，那么当元素很多时，后添加到集合中的元素比较的次数就非常多了。也就是说，如果集合中现在已经有1000个元素，那么第1001个元素加入集合时，它就要调用1000次equals方法。这显然会大大降低效率。

于是，Java采用了哈希表的原理。哈希算法也称为散列算法，是将数据依特定算法直接指定到一个地址上。可以这样简单理解，hashCode方法实际上返回的就是对象存储位置的映像。

当集合中添加元素时，先调用hasCode方法，如果这个位置没有元素，那么就直接存储在这里，如果有元素，那么再调用equals方法与新元素比较，如果相同不存了，如果不相同则说明发生碰撞（冲突），碰撞解决方法多样，最终都会存储在一个合适的位置。有了hasCode后，调用equals的次数大大降低，一般一两次就搞定了。

扩展的内容有兴趣可以自行深入学习。为什么系列采用简短，易懂方式归纳总结一些工作中、面试中常常出现的问题进行解答。如果有错误请及时联系我，以免误导他人。