前言
大家都知道，equals和hashcode是java.lang.object類的兩個重要的方法，在實際應用中常常需要重寫這兩個方法，但至于為什么重寫這兩個方法很多人都搞不明白。
在上一篇博文java中equals和==的區別中介紹了object類的equals方法，并且也介紹了我們可在重寫equals方法，本章我們來說一下為什么重寫equals方法的時候也要重寫hashcode方法。
先讓我們來看看object類源碼
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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：是一個native方法，返回的是對象的內存地址，
equals：對于基本數據類型，==比較的是兩個變量的值。對于引用對象，==比較的是兩個對象的地址。
接下來我們看下hashcode的注釋
1.在 java 應用程序執行期間，在對同一對象多次調用 hashcode 方法時，必須一致地返回相同的整數，前提是將對象進行 equals 比較時所用的信息沒有被修改。
從某一應用程序的一次執行到同一應用程序的另一次執行，該整數無需保持一致。
2.如果根據 equals(object) 方法，兩個對象是相等的，那么對這兩個對象中的每個對象調用 hashcode 方法都必須生成相同的整數結果。
3.如果根據 equals(java.lang.object) 方法，兩個對象不相等，那么兩個對象不一定必須產生不同的整數結果。
但是，程序員應該意識到，為不相等的對象生成不同整數結果可以提高哈希表的性能。
從hashcode的注釋中我們看到，hashcode方法在定義時做出了一些常規協定，即
1，當obj1.equals(obj2) 為 true 時，obj1.hashcode() == obj2.hashcode()
2，當obj1.equals(obj2) 為 false 時，obj1.hashcode() != obj2.hashcode()
hashcode是用于散列數據的快速存取，如利用hashset/hashmap/hashtable類來存儲數據時，都是根據存儲對象的hashcode值來進行判斷是否相同的。如果我們將對象的equals方法重寫而不重寫hashcode，當我們再次new一個新的對象的時候，equals方法返回的是true，但是hashcode方法返回的就不一樣了，如果需要將這些對象存儲到結合中（比如：set，map ...）的時候就違背了原有集合的原則，下面讓我們通過一段代碼看下。
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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方法的都不重寫
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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:356573597p1的hashcode:1735600054falsefalsecom.blueskyli.練習.person@677327b6com.blueskyli.練習.person@1540e19d |
只重寫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:356573597p1的hashcode:1735600054truefalsecom.blueskyli.練習.person@677327b6com.blueskyli.練習.person@1540e19d |
equals和hashcode方法都重寫
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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:3254239p1的hashcode:3254239truefalsecom.blueskyli.練習.person@31a7df |
我們知道map是不允許存在相同的key的，由上面的代碼可以知道，如果不重寫equals和hashcode方法的話會使得你在使用map的時候出現與預期不一樣的結果，具體equals和hashcode如何重寫，里面的邏輯如何實現需要根據現實當中的業務來規定。
總結：
1，兩個對象，用==比較比較的是地址，需采用equals方法（可根據需求重寫）比較。
2，重寫equals()方法就重寫hashcode()方法。
3，一般相等的對象都規定有相同的hashcode。
4，string類重寫了equals和hashcode方法，比較的是值。
5，重寫hashcode方法為了將數據存入hashset/hashmap/hashtable（可以參考源碼有助于理解）類時進行比較
好了，以上就是這篇文章的全部內容了，希望本文的內容對大家的學習或者工作具有一定的參考學習價值，如果有疑問大家可以留言交流，謝謝大家對服務器之家的支持。
原文鏈接：https://www.cnblogs.com/blueskyli/p/9936076.html
