转自https://www.cnblogs.com/keyi/p/7119825.html
1、默认情况(没有覆盖equals方法)下equals方法都是调用Object类的equals方法,而Object的equals方法主要用于判断对象的内存地址引用是不是同一个地址(是不是同一个对象)。
2 、要是类中覆盖了equals方法,那么就要根据具体的代码来确定equals方法的作用了,覆盖后一般都是通过对象的内容是否相等来判断对象是否相等。
没有覆盖equals方法代码如下:
- //学生类
- public class Student {
- private int age;
- private String name;
- public Student() {
- }
- public Student(int age, String name) {
- super();
- this.age = age;
- this.name = name;
- }
- public int getAge() {
- return age;
- }
- public String getName() {
- return name;
- }
- public void setAge(int age) {
- this.age = age;
- }
- public void setName(String name) {
- this.name = name;
- }
- }
- import java.util.HashSet;
- import java.util.LinkedList;
- import java.util.Set;
- public class EqualsTest {
- public static void main(String[] args) {
- LinkedList<Student> list = new LinkedList<Student>();
- Set<Student> set = new HashSet<Student>();
- Student stu1 = new Student(3,"张三");
- Student stu2 = new Student(3,"张三");
- System.out.println("stu1 == stu2 : "+(stu1 == stu2));
- System.out.println("stu1.equals(stu2) : "+stu1.equals(stu2));
- list.add(stu1);
- list.add(stu2);
- System.out.println("list size:"+ list.size());
- set.add(stu1);
- set.add(stu2);
- System.out.println("set size:"+ set.size());
- }
- }
运行结果:
stu1 == stu2 : falsestu1.equals(stu2) : false
list size:2
set size:2
结果分析:Student类没有覆盖equals方法,stu1调用equals方法实际上调用的是Object的equals方法。所以采用对象内存地址是否相等来判断对象是否相等。因为是两个新对象所以对象的内存地址不相等,所以stu1.equals(stu2) 是false。
3、我们覆盖一下equals方法(age和name属性),让Student类其通过判断对象的内容是否相等来确定对象是否相等。
覆盖后的Student类:
- //学生类
- public class Student {
- private int age;
- private String name;
- public Student() {
- }
- public Student(int age, String name) {
- super();
- this.age = age;
- this.name = name;
- }
- public int getAge() {
- return age;
- }
- public String getName() {
- return name;
- }
- public void setAge(int age) {
- this.age = age;
- }
- public void setName(String name) {
- this.name = name;
- }
- @Override
- public boolean equals(Object obj) {
- if (this == obj)
- return true;
- if (obj == null)
- return false;
- if (getClass() != obj.getClass())
- return false;
- Student other = (Student) obj;
- if (age != other.age)
- return false;
- if (name == null) {
- if (other.name != null)
- return false;
- } else if (!name.equals(other.name))
- return false;
- return true;
- }
- }
stu1 == stu2 : false
stu1.equals(stu2) : true
list size:2
set size:2
结果分析:因为Student两个对象的age和name属性相等,而且又是通过覆盖equals方法来判断的,所示stu1.equals(stu2) 为true。注意以上几次测试list和set的size都是2
二、HashCode
4、通过以上的代码运行,我们知道equals方法已经生效。接下来我们在覆盖一下hashCode方法(通过age和name属性来生成hashcode)并不覆盖equals方法,其中Hash码是通过age和name生成的。
覆盖hashcode后的Student类:
- //学生类
- public class Student {
- private int age;
- private String name;
- public Student() {
- }
- public Student(int age, String name) {
- super();
- this.age = age;
- this.name = name;
- }
- public int getAge() {
- return age;
- }
- public String getName() {
- return name;
- }
- public void setAge(int age) {
- this.age = age;
- }
- public void setName(String name) {
- this.name = name;
- }
- @Override
- public int hashCode() {
- final int prime = 31;
- int result = 1;
- result = prime * result + age;
- result = prime * result + ((name == null) ? 0 : name.hashCode());
- return result;
- }
- }
运行结果:
stu1 == stu2 : false
stu1.equals(stu2) : false
list size:2
hashCode :775943
hashCode :775943
set size:2
结果分析:我们并没有覆盖equals方法只覆盖了hashCode方法,两个对象虽然hashCode一样,但在将stu1和stu2放入set集合时由于equals方法比较的两个对象是false,所以就没有在比较两个对象的hashcode值。
5、我们覆盖一下equals方法和hashCode方法。
Student代码如下:
- //学生类
- public class Student {
- private int age;
- private String name;
- public Student() {
- }
- public Student(int age, String name) {
- super();
- this.age = age;
- this.name = name;
- }
- public int getAge() {
- return age;
- }
- public String getName() {
- return name;
- }
- public void setAge(int age) {
- this.age = age;
- }
- public void setName(String name) {
- this.name = name;
- }
- @Override
- public int hashCode() {
- final int prime = 31;
- int result = 1;
- result = prime * result + age;
- result = prime * result + ((name == null) ? 0 : name.hashCode());
- System.out.println("hashCode : "+ result);
- return result;
- }
- @Override
- public boolean equals(Object obj) {
- if (this == obj)
- return true;
- if (obj == null)
- return false;
- if (getClass() != obj.getClass())
- return false;
- Student other = (Student) obj;
- if (age != other.age)
- return false;
- if (name == null) {
- if (other.name != null)
- return false;
- } else if (!name.equals(other.name))
- return false;
- return true;
- }
- }
运行结果:
stu1 == stu2 : false
stu1.equals(stu2) :true
list size:2
hashCode :775943
hashCode :775943
set size:1
结果分析:stu1和stu2通过equals方法比较相等,而且返回的hashCode值一样,所以放入set集合中时只放入了一个对象。
6、下面我们让两个对象equals方法比较相等,但hashCode值不相等试试。
Student类的代码如下:
- //学生类
- public class Student {
- private int age;
- private String name;
- <span style="color:#ff0000;">private static int index=5;</span>
- public Student() {
- }
- public Student(int age, String name) {
- super();
- this.age = age;
- this.name = name;
- }
- public int getAge() {
- return age;
- }
- public String getName() {
- return name;
- }
- public void setAge(int age) {
- this.age = age;
- }
- public void setName(String name) {
- this.name = name;
- }
- @Override
- public int hashCode() {
- final int prime = 31;
- int result = 1;
- result = prime * result + <span style="color:#ff0000;">(age+index++)</span>;
- result = prime * result + ((name == null) ? 0 : name.hashCode());
- <span style="color:#ff0000;">System.out.println("result :"+result);</span>
- return result;
- }
- @Override
- public boolean equals(Object obj) {
- if (this == obj)
- return true;
- if (obj == null)
- return false;
- if (getClass() != obj.getClass())
- return false;
- Student other = (Student) obj;
- if (age != other.age)
- return false;
- if (name == null) {
- if (other.name != null)
- return false;
- } else if (!name.equals(other.name))
- return false;
- return true;
- }
- }
stu1 == stu2 : false
stu1.equals(stu2) : true
list size:2
hashCode :776098
hashCode :776129
set size:2
结果分析:虽然stu1和stu2通过equals方法比较相等,但两个对象的hashcode的值并不相等,所以在将stu1和stu2放入set集合中时认为是两个不同的对象。
7、修改stu1的某个属性值
Student代码如下:
- //学生类
- public class Student {
- private int age;
- private String name;
- public Student() {
- }
- public Student(int age, String name) {
- super();
- this.age = age;
- this.name = name;
- }
- public int getAge() {
- return age;
- }
- public String getName() {
- return name;
- }
- public void setAge(int age) {
- this.age = age;
- }
- public void setName(String name) {
- this.name = name;
- }
- @Override
- public int hashCode() {
- final int prime = 31;
- int result = 1;
- result = prime * result + age;
- result = prime * result + ((name == null) ? 0 : name.hashCode());
- System.out.println("hashCode : "+ result);
- return result;
- }
- @Override
- public boolean equals(Object obj) {
- if (this == obj)
- return true;
- if (obj == null)
- return false;
- if (getClass() != obj.getClass())
- return false;
- Student other = (Student) obj;
- if (age != other.age)
- return false;
- if (name == null) {
- if (other.name != null)
- return false;
- } else if (!name.equals(other.name))
- return false;
- return true;
- }
- }
测试代码如下:
- import java.util.HashSet;
- import java.util.LinkedList;
- import java.util.Set;
- public class EqualsTest {
- public static void main(String[] args) {
- LinkedList<Student> list = new LinkedList<Student>();
- Set<Student> set = new HashSet<Student>();
- Student stu1 = new Student(3,"张三");
- Student stu2 = new Student(3,"张三");
- System.out.println("stu1 == stu2 : "+(stu1 == stu2));
- System.out.println("stu1.equals(stu2) : "+stu1.equals(stu2));
- list.add(stu1);
- list.add(stu2);
- System.out.println("list size:"+ list.size());
- set.add(stu1);
- set.add(stu2);
- System.out.println("set size:"+ set.size());
- stu1.setAge(34);
- System.out.println("remove stu1 : "+set.remove(stu1));
- System.out.println("set size:"+ set.size());
- }
- }
stu1 == stu2 : false
stu1.equals(stu2) : true
list size:2
hashCode : 775943
hashCode : 775943
set size:1
hashCode : 776904
remove stu1 : false
set size:1
结果分析:
当我们将某个对象存到set中时,如果该对象的属性参与了hashcode的计算,那么以后就不能修改该对象参与hashcode计算的那些属性了,否则会引起意向不到的错误的。正如测试中,不能够移除stu1对象。
总结:
1、equals方法用于比较对象的内容是否相等(覆盖以后)
2、hashcode方法只有在集合中用到
3、当覆盖了equals方法时,比较对象是否相等将通过覆盖后的equals方法进行比较(判断对象的内容是否相等)。
4、将对象放入到集合中时,首先判断要放入对象的hashcode值与集合中的任意一个元素的hashcode值是否相等,如果不相等直接将该对象放入集合中。如果hashcode值相等,然后再通过equals方法判断要放入对象与集合中的任意一个对象是否相等,如果equals判断不相等,直接将该元素放入到集合中,否则不放入。
5、将元素放入集合的流程图:
6、HashSet中add方法源代码:
- public boolean add(E e) {
- return map.put(e, PRESENT)==null;
- }
- <pre name="code" class="java"> public V put(K key, V value) {
- if (key == null)
- return putForNullKey(value);
- int hash = hash(key.hashCode());
- int i = indexFor(hash, table.length);
- for (Entry<K,V> e = table[i]; e != null; e = e.next) {
- Object k;
- if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
- V oldValue = e.value;
- e.value = value;
- e.recordAccess(this);
- return oldValue;
- }
- }
- modCount++;
- addEntry(hash, key, value, i);
- return null;
- }</pre>
- <pre></pre>
- <pre></pre>