hashcode的作用

      在Hashtable，HashMap，HashSet中, 根据hashcode来决定是否两个对象是否相同，如果相同的话则会覆盖已经存在的对象。

      所以创建类的时候，如果重写了equals方法，记得一定也要重写hashcode方法。因为在Hashtable,HashMap,HashSet中并不是根据equals来决定对象是否是同一个，而是根据hashcode。

      基本原则是：如果两个对象是equal的，那么要保证他们的hashcode也是一致的。

      除此之外hashcode并无多大作用。

 

 举例：Student类 , equals和hashcode方法都重写了。

 

package hashcode;

import java.util.HashMap;
import java.util.HashSet;
import java.util.Hashtable;

public class Student {

	private int age;
	private String name;

	public int getAge() {
		return age;
	}

	public void setAge(int age) {
		this.age = age;
	}

	public String getName() {
		return name;
	}

	public void setName(String name) {
		this.name = name;
	}

	/**
	 * hashcode is used in hashset, hashmap, hashtable. if key's hashcode is the
	 * same, the key will be overided.
	 */
	public int hashCode() {

		return this.age + this.name.hashCode();
	}

	public boolean equals(Object obj) {
		if (obj == null || !(obj instanceof Student)) {
			return false;
		}
		if (this == obj) {
			return true;
		}
		Student s = (Student) obj;
		if (s.getAge() == this.getAge() && s.getName().equals(this.getName())) {
			return true;
		} else {
			return false;
		}
	}

	public Student() {
	}

	public Student(String name, int age) {
		setName(name);
		setAge(age);
	}

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		Student s1 = new Student("xiaohong", 11);
		Student s2 = new Student("xiaohong", 11);
		System.out.println(s1.hashCode()); //265195626
		System.out.println(s2.hashCode()); //265195626 
		if (s1.equals(s2)) {
			System.out.println("s1 equals s2");//This line prints.
		} else {
			System.out.println("s1 doesn't equal s2");
		}
		HashMap<Student, String> map = new HashMap<Student, String>();
		map.put(s1, "ok");
		map.put(s2, "nok");

		System.out.println(map.get(s1));//"nok", s1 is overided by s2.
		System.out.println(map.get(s2));//"nok"

		Hashtable<Student, String> table = new Hashtable<Student, String>();
		table.put(s1, "ok");
		table.put(s2, "nok");

		System.out.println(map.get(s1));//"nok", s1 is overided by s2.
		System.out.println(map.get(s2));//"nok"

		HashSet<Student> set = new HashSet<Student>();
		Student[] setArray = new Student[2];
		set.add(s1);
		set.add(s2);

		for (Student s : set.toArray(setArray)) {
			if (s != null)
				System.out.println(s.getName() + s.getAge());

		} // print only one element. s1 is overied by s2.

	}

}

 假如只重写equals方法，而不重写hashcode方法，那么在hashtable，hashSet，hashMap中，这两个本来是equal的Student对象也会被认为是不同的对象而分别存放。