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java中為何重寫equals時必須重寫hashCode方法詳解

2024-07-14 08:42:50
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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

/** * 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。

/** * @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

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:356573597p1的hashCode:1735600054falsefalsecom.blueskyli.練習[email protected].練習.Person@1540e19d

只重寫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:356573597p1的hashCode:1735600054truefalsecom.blueskyli.練習[email protected].練習.Person@1540e19d

equals和hashCode方法都重寫

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: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VeVb武林網的支持。


注:相關教程知識閱讀請移步到JAVA教程頻道。
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