java中为什么重写equals时必须重写hashCode方法
小编给大家分享一下java中为什么重写equals时必须重写hashCode方法，相信大部分人都还不怎么了解，因此分享这篇文章给大家参考一下，希望大家阅读完这篇文章后大有收获，下面让我们一起去了解一下吧！
创新互联公司一直在为企业提供服务，多年的磨炼，使我们在创意设计，成都营销网站建设到技术研发拥有了开发经验。我们擅长倾听企业需求，挖掘用户对产品需求服务价值，为企业制作有用的创意设计体验。核心团队拥有超过十年以上行业经验，涵盖创意，策化，开发等专业领域，公司涉及领域有基础互联网服务达州托管服务器、app软件开发公司、手机移动建站、网页设计、网络整合营销。
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}.
*
* The general contract of {@code hashCode} is:
*
-
*
- 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. *
- 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. *
- It is not 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. *
* 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. ** The {@code equals} method implements an equivalence relation * on non-null object references: *
-
*
- It is reflexive: for any non-null reference value * {@code x}, {@code x.equals(x)} should return * {@code true}. *
- It is symmetric: 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}. *
- It is transitive: 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}. *
- It is consistent: 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. *
- For any non-null reference value {@code x}, * {@code x.equals(null)} should return {@code false}. *
* 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}). *
* 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 map = new HashMap();
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 com.blueskyli.练习.Person@1540e19d
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 com.blueskyli.练习.Person@1540e19d
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
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
以上是“java中为什么重写equals时必须重写hashCode方法”这篇文章的所有内容，感谢各位的阅读！相信大家都有了一定的了解，希望分享的内容对大家有所帮助，如果还想学习更多知识，欢迎关注创新互联行业资讯频道！
网站题目：java中为什么重写equals时必须重写hashCode方法
文章转载：http://6mz.cn/article/ispocs.html