java中為何重寫equals時必須重寫hashCode方法詳解
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(qū)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(shù)據(jù)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(zhí)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(shù),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(zhí)行到同一應用程序的另一次執(zhí)行,該整數(shù)無需保持一致。
2.如果根據(jù) 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(shù)結果。
3.如果根據(jù) 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(chǎn)生不同的整數(shù)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(shù)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(guī)協(xié)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(shù)據(jù)的快速存取,如利用HashSet/HashMap/Hashtable類來存儲數(shù)據(jù)時,都是根據(jù)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:356573597 p1的hashCode:1735600054 false false com.blueskyli.練習.Person@677327b6 com.blueskyli.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:356573597 p1的hashCode:1735600054 true false com.blueskyli.練習.Person@677327b6 com.blueskyli.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:3254239 p1的hashCode:3254239 true false com.blueskyli.練習.Person@31a7df
我們知道m(xù)ap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(xiàn)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(xiàn)需要根據(jù)現(xiàn)實當中的業(yè)務來規(guī)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(jù)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(guī)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(shù)據(jù)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,謝謝大家對腳本之家的支持。
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