String类和包装器类的equals()和hashCode()方法实现与Object类不同。Object类的equals()方法比较对象的引用，而不是内容。Object类的hashCode()方法为每个对象返回不同的hashCode，无论内容是否相同。

It leads problem when you use Map collection and the key is of Persistent type, StringBuffer/builder type. Since they don't override equals() and hashCode() unlike String class, equals() will return false when you compare two different objects even though both have same contents. It will make the hashMap storing same content keys. Storing same content keys means it is violating the rule of Map because Map doesnt allow duplicate keys at all. Therefore you override equals() as well as hashCode() methods in your class and provide the implementation(IDE can generate these methods) so that they work same as String's equals() and hashCode() and prevent same content keys.

你必须重写hashCode()方法和equals()，因为equals()根据hashCode工作。

此外，与equals()一起重写hashCode()方法有助于完好无损equals()-hashCode()契约:“如果两个对象相等，那么它们必须具有相同的哈希码。”

什么时候需要为hashCode()编写自定义实现?

正如我们所知，HashMap的内部工作是基于哈希原理的。条目集存储在特定的存储桶中。您可以根据自己的需求定制hashCode()实现，以便相同的类别对象可以存储在相同的索引中。 当你使用put(k,v)方法将值存储到Map集合中时，put()的内部实现是:

put(k, v){
hash(k);
index=hash & (n-1);
}

意思是，它生成索引，索引是基于特定键对象的hashcode生成的。所以让这个方法根据你的需求生成哈希码，因为相同的哈希码条目集将存储在相同的桶或索引中。

就是这样!

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.

因为如果你不重写它们，你将使用Object中的默认实现。

考虑到实例相等和hascode值通常需要了解组成对象的内容，它们通常需要在类中重新定义，以具有任何有形的意义。

简单地说，Object中的equals-方法检查引用是否相等，当属性相等时，类的两个实例在语义上仍然相等。例如，当将对象放入使用等号和hashcode(如HashMap和Set)的容器中时，这很重要。假设我们有这样一个类:

public class Foo {
    String id;
    String whatevs;

    Foo(String id, String whatevs) {
        this.id = id;
        this.whatevs = whatevs;
    }
}

我们创建了两个具有相同id的实例:

Foo a = new Foo("id", "something");
Foo b = new Foo("id", "something else");

如果不重写等号，我们将得到:

A.equals (b)为假，因为它们是两个不同的实例 a.equals(a)为真，因为它是同一个实例 b.equals(b)为真，因为它是同一个实例

正确吗?也许吧，如果这是你想要的。但假设我们希望具有相同id的对象是相同的对象，不管它是否是两个不同的实例。我们重写等号(和hashcode):

public class Foo {
    String id;
    String whatevs;

    Foo(String id, String whatevs) {
        this.id = id;
        this.whatevs = whatevs;
    }

    @Override
    public boolean equals(Object other) {
        if (other instanceof Foo) {
            return ((Foo)other).id.equals(this.id);   
        }
    }

    @Override
    public int hashCode() {
        return this.id.hashCode();
    }
}

至于实现equals和hashcode，我建议使用Guava的helper方法

Bah -“你必须在每个重写equals()的类中重写hashCode()。”

[出自Joshua Bloch的《Effective Java》?］

Isn't this the wrong way round? Overriding hashCode likely implies you're writing a hash-key class, but overriding equals certainly does not. There are many classes that are not used as hash-keys, but do want a logical-equality-testing method for some other reason. If you choose "equals" for it, you may then be mandated to write a hashCode implementation by overzealous application of this rule. All that achieves is adding untested code in the codebase, an evil waiting to trip someone up in the future. Also writing code you don't need is anti-agile. It's just wrong (and an ide generated one will probably be incompatible with your hand-crafted equals).

他们肯定应该在被写来用作键的对象上强制设置一个接口吗?无论如何，Object永远不应该提供默认的hashCode()和equals() imho。它可能鼓励了许多破碎的散列集合。

但无论如何，我认为“规则”是前后颠倒的。与此同时，我将继续避免使用“等号”进行相等性测试方法:-(

当您希望将自定义对象作为Map中的键存储和检索时，您应该始终重写自定义对象中的equals和hashCode。 例如:

Person p1 = new Person("A",23);
Person p2 = new Person("A",23);
HashMap map = new HashMap();
map.put(p1,"value 1");
map.put(p2,"value 2");

这里p1和p2只被认为是一个对象，映射大小也只有1，因为它们相等。