在上一篇博文Java中equals和==的区别中介绍了Object类的equals方法，并且也介绍了我们可在重写equals方法，本章我们来说一下为什么重写equals方法的时候也要重写hashCode方法。
先让我们来看看Object类源码
/** * 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); }
hashCode：是一个native方法，返回的是对象的内存地址，
equals：对于基本数据类型，==比较的是两个变量的值。对于引用对象，==比较的是两个对象的地址。
接下来我们看下hashCode的注释
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，那么两个对象不一定必须产生不同的整数结果。
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。
从hashCode的注释中我们看到，hashCode方法在定义时做出了一些常规协定，即
1，当obj1.equals(obj2) 为 true 时，obj1.hashCode() == obj2.hashCode()
2，当obj1.equals(obj2) 为 false 时，obj1.hashCode() != obj2.hashCode()
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。如果我们将对象的equals方法重写而不重写hashcode，当我们再次new一个新的对象的时候，equals方法返回的是true，但是hashCode方法返回的就不一样了，如果需要将这些对象存储到结合中（比如：Set，Map ...）的时候就违背了原有集合的原则，下面让我们通过一段代码看下。
/** * @see Person * @param args */ public static void main(String[] args) { HashMap<Person, Integer> map = new HashMap<Person, Integer>(); Person p = new Person("jack",22,"男"); Person p1 = new Person("jack",22,"男"); System.out.println("p的hashCode:"+p.hashCode()); System.out.println("p1的hashCode:"+p1.hashCode()); System.out.println(p.equals(p1)); System.out.println(p == p1); map.put(p,888); map.put(p1,888); map.forEach((key,val)->{ System.out.println(key); System.out.println(val); }); }
equals和hashCode方法的都不重写
public class Person { private String name; private int age; private String sex; Person(String name,int age,String sex){ this.name = name; this.age = age; this.sex = sex; } }
p的hashCode:356573597 p1的hashCode:1735600054 false false com.blueskyli.练习.Person@677327b6 888 com.blueskyli.练习.Person@1540e19d 888
只重写equals方法
public class Person { private String name; private int age; private String sex; Person(String name,int age,String sex){ this.name = name; this.age = age; this.sex = sex; } @Override public boolean equals(Object obj) { if(obj instanceof Person){ Person person = (Person)obj; return name.equals(person.name); } return super.equals(obj); } }
p的hashCode:356573597 p1的hashCode:1735600054 true false com.blueskyli.练习.Person@677327b6 888 com.blueskyli.练习.Person@1540e19d 888
equals和hashCode方法都重写
public class Person { private String name; private int age; private String sex; Person(String name,int age,String sex){ this.name = name; this.age = age; this.sex = sex; } @Override public boolean equals(Object obj) { if(obj instanceof Person){ Person person = (Person)obj; return name.equals(person.name); } return super.equals(obj); } @Override public int hashCode() { return name.hashCode(); } }
p的hashCode:3254239 p1的hashCode:3254239 true false com.blueskyli.练习.Person@31a7df 888
我们知道map是不允许存在相同的key的，由上面的代码可以知道，如果不重写equals和hashCode方法的话会使得你在使用map的时候出现与预期不一样的结果，具体equals和hashCode如何重写，里面的逻辑如何实现需要根据现实当中的业务来规定。
总结：
1，两个对象，用==比较比较的是地址，需采用equals方法（可根据需求重写）比较。
2，重写equals()方法就重写hashCode()方法。
3，一般相等的对象都规定有相同的hashCode。
4，String类重写了equals和hashCode方法，比较的是值。
5，重写hashcode方法为了将数据存入HashSet/HashMap/Hashtable（可以参考源码有助于理解）类时进行比较