1、默認情況(沒有覆蓋equals方法)下equals方法都是調用Object類的equals方法,而Object的equals方法主要用於判斷對象的內存地址引用是不是同一個地址(是不是同一個對象)。
2 、要是類中覆蓋了equals方法,那麼就要根據具體的代碼來確定equals方法的作用了,覆蓋後一般都是通過對象的內容是否相等來判斷對象是否相等。
沒有覆蓋equals方法代碼如下:
- //學生類
- 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,"張三");
- Student stu2 = new Student(3,"張三");
- 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());
- }
- }
運行結果:
stu1 == stu2 : falsestu1.equals(stu2) : false
list size:2
set size:2
結果分析:Student類沒有覆蓋equals方法,stu1調用equals方法實際上調用的是Object的equals方法。所以採用對象內存地址是否相等來判斷對象是否相等。因爲是兩個新對象所以對象的內存地址不相等,所以stu1.equals(stu2) 是false。
3、我們覆蓋一下equals方法(age和name屬性),讓Student類其通過判斷對象的內容是否相等來確定對象是否相等。
覆蓋後的Student類:
- //學生類
- 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;
- }
- }
運行結果:
stu1 == stu2 : false
stu1.equals(stu2) : true
list size:2
set size:2
結果分析:因爲Student兩個對象的age和name屬性相等,而且又是通過覆蓋equals方法來判斷的,所示stu1.equals(stu2) 爲true。注意以上幾次測試list和set的size都是2
二、HashCode
4、通過以上的代碼運行,我們知道equals方法已經生效。接下來我們在覆蓋一下hashCode方法(通過age和name屬性來生成hashcode)並不覆蓋equals方法,其中Hash碼是通過age和name生成的。
覆蓋hashcode後的Student類:
- //學生類
- 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;
- }
- }
運行結果:
stu1 == stu2 : false
stu1.equals(stu2) : false
list size:2
hashCode :775943
hashCode :775943
set size:2
結果分析:我們並沒有覆蓋equals方法只覆蓋了hashCode方法,兩個對象雖然hashCode一樣,但在將stu1和stu2放入set集合時由於equals方法比較的兩個對象是false,所以就沒有在比較兩個對象的hashcode值。
5、我們覆蓋一下equals方法和hashCode方法。
Student代碼如下:
- //學生類
- 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;
- }
- }
運行結果:
stu1 == stu2 : false
stu1.equals(stu2) :true
list size:2
hashCode :775943
hashCode :775943
set size:1
結果分析:stu1和stu2通過equals方法比較相等,而且返回的hashCode值一樣,所以放入set集合中時只放入了一個對象。
6、下面我們讓兩個對象equals方法比較相等,但hashCode值不相等試試。
Student類的代碼如下:
- //學生類
- 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
結果分析:雖然stu1和stu2通過equals方法比較相等,但兩個對象的hashcode的值並不相等,所以在將stu1和stu2放入set集合中時認爲是兩個不同的對象。
7、修改stu1的某個屬性值
Student代碼如下:
- //學生類
- 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;
- }
- }
測試代碼如下:
- 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,"張三");
- Student stu2 = new Student(3,"張三");
- 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());
- }
- }
運行結果:
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
結果分析:
當我們將某個對象存到set中時,如果該對象的屬性參與了hashcode的計算,那麼以後就不能修改該對象參與hashcode計算的那些屬性了,否則會引起意向不到的錯誤的。正如測試中,不能夠移除stu1對象。
總結:
1、equals方法用於比較對象的內容是否相等(覆蓋以後)
2、hashcode方法只有在集合中用到
3、當覆蓋了equals方法時,比較對象是否相等將通過覆蓋後的equals方法進行比較(判斷對象的內容是否相等)。
4、將對象放入到集合中時,首先判斷要放入對象的hashcode值與集合中的任意一個元素的hashcode值是否相等,如果不相等直接將該對象放入集合中。如果hashcode值相等,然後再通過equals方法判斷要放入對象與集合中的任意一個對象是否相等,如果equals判斷不相等,直接將該元素放入到集合中,否則不放入。
5、將元素放入集合的流程圖:
6、HashSet中add方法源代碼:
- 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>
- 原文:http://blog.csdn.net/afgasdg/article/details/6889383