最近我读了这篇文章 开发人员工作文件。
该文档是关于有效和正确地定义hashCode()和equals()，但我无法弄清楚为什么我们需要覆盖这两个方法。
我如何决定有效地实现这些方法?
当前回答
HashMap和HashSet等集合使用对象的hashcode值来确定该对象应该如何存储在集合中，然后再次使用hashcode来定位该对象 在它的收藏中。
哈希检索是一个两步过程:
找到正确的桶(使用hashCode()) 在桶中搜索正确的元素(使用equals())
下面是一个关于为什么我们应该重写equals()和hashcode()的小例子。
考虑一个Employee类，它有两个字段:年龄和名字。
public class Employee {
String name;
int age;
public Employee(String name, int age) {
this.name = name;
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
@Override
public boolean equals(Object obj) {
if (obj == this)
return true;
if (!(obj instanceof Employee))
return false;
Employee employee = (Employee) obj;
return employee.getAge() == this.getAge()
&& employee.getName() == this.getName();
}
// commented
/* @Override
public int hashCode() {
int result=17;
result=31*result+age;
result=31*result+(name!=null ? name.hashCode():0);
return result;
}
*/
}
现在创建一个类，将Employee对象插入到HashSet中并测试该对象是否存在。
public class ClientTest {
public static void main(String[] args) {
Employee employee = new Employee("rajeev", 24);
Employee employee1 = new Employee("rajeev", 25);
Employee employee2 = new Employee("rajeev", 24);
HashSet<Employee> employees = new HashSet<Employee>();
employees.add(employee);
System.out.println(employees.contains(employee2));
System.out.println("employee.hashCode(): " + employee.hashCode()
+ " employee2.hashCode():" + employee2.hashCode());
}
}
它将打印以下内容:
false
employee.hashCode(): 321755204 employee2.hashCode():375890482
现在uncomment hashcode()方法，执行相同的方法，输出将是:
true
employee.hashCode(): -938387308 employee2.hashCode():-938387308
Now can you see why if two objects are considered equal, their hashcodes must also be equal? Otherwise, you'd never be able to find the object since the default hashcode method in class Object virtually always comes up with a unique number for each object, even if the equals() method is overridden in such a way that two or more objects are considered equal. It doesn't matter how equal the objects are if their hashcodes don't reflect that. So one more time: If two objects are equal, their hashcodes must be equal as well.
其他回答
在下面的例子中，如果您注释掉Person类中equals或hashcode的覆盖，此代码将无法查找Tom的订单。使用哈希码的默认实现可能会导致哈希表查找失败。
下面是一个简化的代码，它按Person提取人们的订单。Person被用作哈希表中的键。
public class Person {
String name;
int age;
String socialSecurityNumber;
public Person(String name, int age, String socialSecurityNumber) {
this.name = name;
this.age = age;
this.socialSecurityNumber = socialSecurityNumber;
}
@Override
public boolean equals(Object p) {
//Person is same if social security number is same
if ((p instanceof Person) && this.socialSecurityNumber.equals(((Person) p).socialSecurityNumber)) {
return true;
} else {
return false;
}
}
@Override
public int hashCode() { //I am using a hashing function in String.java instead of writing my own.
return socialSecurityNumber.hashCode();
}
}
public class Order {
String[] items;
public void insertOrder(String[] items)
{
this.items=items;
}
}
import java.util.Hashtable;
public class Main {
public static void main(String[] args) {
Person p1=new Person("Tom",32,"548-56-4412");
Person p2=new Person("Jerry",60,"456-74-4125");
Person p3=new Person("Sherry",38,"418-55-1235");
Order order1=new Order();
order1.insertOrder(new String[]{"mouse","car charger"});
Order order2=new Order();
order2.insertOrder(new String[]{"Multi vitamin"});
Order order3=new Order();
order3.insertOrder(new String[]{"handbag", "iPod"});
Hashtable<Person,Order> hashtable=new Hashtable<Person,Order>();
hashtable.put(p1,order1);
hashtable.put(p2,order2);
hashtable.put(p3,order3);
//The line below will fail if Person class does not override hashCode()
Order tomOrder= hashtable.get(new Person("Tom", 32, "548-56-4412"));
for(String item:tomOrder.items)
{
System.out.println(item);
}
}
}
HashMap和HashSet等集合使用对象的hashcode值来确定该对象应该如何存储在集合中，然后再次使用hashcode来定位该对象 在它的收藏中。
哈希检索是一个两步过程:
找到正确的桶(使用hashCode()) 在桶中搜索正确的元素(使用equals())
下面是一个关于为什么我们应该重写equals()和hashcode()的小例子。
考虑一个Employee类，它有两个字段:年龄和名字。
public class Employee {
String name;
int age;
public Employee(String name, int age) {
this.name = name;
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
@Override
public boolean equals(Object obj) {
if (obj == this)
return true;
if (!(obj instanceof Employee))
return false;
Employee employee = (Employee) obj;
return employee.getAge() == this.getAge()
&& employee.getName() == this.getName();
}
// commented
/* @Override
public int hashCode() {
int result=17;
result=31*result+age;
result=31*result+(name!=null ? name.hashCode():0);
return result;
}
*/
}
现在创建一个类，将Employee对象插入到HashSet中并测试该对象是否存在。
public class ClientTest {
public static void main(String[] args) {
Employee employee = new Employee("rajeev", 24);
Employee employee1 = new Employee("rajeev", 25);
Employee employee2 = new Employee("rajeev", 24);
HashSet<Employee> employees = new HashSet<Employee>();
employees.add(employee);
System.out.println(employees.contains(employee2));
System.out.println("employee.hashCode(): " + employee.hashCode()
+ " employee2.hashCode():" + employee2.hashCode());
}
}
它将打印以下内容:
false
employee.hashCode(): 321755204 employee2.hashCode():375890482
现在uncomment hashcode()方法，执行相同的方法，输出将是:
true
employee.hashCode(): -938387308 employee2.hashCode():-938387308
Now can you see why if two objects are considered equal, their hashcodes must also be equal? Otherwise, you'd never be able to find the object since the default hashcode method in class Object virtually always comes up with a unique number for each object, even if the equals() method is overridden in such a way that two or more objects are considered equal. It doesn't matter how equal the objects are if their hashcodes don't reflect that. So one more time: If two objects are equal, their hashcodes must be equal as well.
如果重写equals()而不是hashcode()，则不会发现任何问题，除非您或其他人在HashSet等散列集合中使用该类类型。 在我之前的人已经清楚地解释了很多次文献理论，我只是在这里提供一个非常简单的例子。
考虑一个类，它的equals()需要表示自定义的东西:-
public class Rishav {
private String rshv;
public Rishav(String rshv) {
this.rshv = rshv;
}
/**
* @return the rshv
*/
public String getRshv() {
return rshv;
}
/**
* @param rshv the rshv to set
*/
public void setRshv(String rshv) {
this.rshv = rshv;
}
@Override
public boolean equals(Object obj) {
if (obj instanceof Rishav) {
obj = (Rishav) obj;
if (this.rshv.equals(((Rishav) obj).getRshv())) {
return true;
} else {
return false;
}
} else {
return false;
}
}
@Override
public int hashCode() {
return rshv.hashCode();
}
}
现在考虑这个主类:-
import java.util.HashSet;
import java.util.Set;
public class TestRishav {
public static void main(String[] args) {
Rishav rA = new Rishav("rishav");
Rishav rB = new Rishav("rishav");
System.out.println(rA.equals(rB));
System.out.println("-----------------------------------");
Set<Rishav> hashed = new HashSet<>();
hashed.add(rA);
System.out.println(hashed.contains(rB));
System.out.println("-----------------------------------");
hashed.add(rB);
System.out.println(hashed.size());
}
}
这将产生以下输出:-
true
-----------------------------------
true
-----------------------------------
1
我对结果很满意。但是如果我没有覆盖hashCode()，它将导致噩梦，因为具有相同成员内容的Rishav对象将不再被视为唯一的hashCode将是不同的，因为由默认行为生成，这里将是输出:-
true
-----------------------------------
false
-----------------------------------
2
考虑在一个桶中收集所有黑色的球。你的工作是像下面这样给这些球上色，并将其用于适当的游戏，
对于网球-黄色，红色。 板球-白色
现在水桶有三种颜色的球黄色，红色和白色。只有你知道哪个颜色适合哪个游戏。
给球上色-哈希。 选择比赛的球-平等。
如果你给球上色，然后有人选了板球或网球，他们不会介意颜色的!!
我正在研究解释“如果你只覆盖hashCode，那么当你调用myMap.put(first,someValue)时，它首先接受，计算它的hashCode并将其存储在给定的桶中。然后，当你调用myMap.put(first,someOtherValue)时，它应该根据Map文档将first替换为second，因为它们是相等的(根据我们的定义)。”：
我认为第二次添加myMap时应该是第二个对象比如myMap。put(second,someOtherValue)
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