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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
|
/**
* 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();
|
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
|
/**
* 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。
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
|
/**
* @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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
|
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;
}
}
|
|
1
2
3
4
5
6
|
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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
|
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);
}
}
|
|
1
2
3
4
5
6
|
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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
|
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();
}
}
|
|
1
2
3
4
5
|
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，谢谢大家对服务器之家的支持。
原文链接：https://www.cnblogs.com/blueskyli/p/9936076.html