前言
大家都知道， equals 和hashcode是 java .lang.object类的两个重要的方法，在实际应用中常常需要 重写 这两个方法，但至于为什么重写这两个方法很多人都搞不明白。
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
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/** * 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(); |
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/** * 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。
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/** * @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
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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; } } |
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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
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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); } } |
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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
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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(); } } |
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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
好了，以上就是这篇文章的全部内容了，希望本文的内容对大家的学习或者工作具有一定的参考学习价值，如果有疑问大家可以留言交流，谢谢大家对的支持。
原文链接：https://www.cnblogs.com/blueskyli/p/9936076.html
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