HashMap,你是怎么做到的Key重复？
小朱: 今天遇到了一件奇怪的事?
大牛: 什么事?
小朱: 我把对象当作Key放到HashMap里,Key重复了?
大牛: Key重复了?出现了两个一样的Key?
小朱: 是啊!
大牛: 有意思,你要是取值,他会返给你哪个?
小朱: 这个没试过.
大牛: 让我看看你的代码.
小朱: 好的.
代码
Student
public class Student {
private Integer id;
private String name;
//get和set省略
public Student(Integer id, String name) {
this.id = id;
this.name = name;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Student student = (Student) o;
return Objects.equals(id, student.id) &&
Objects.equals(name, student.name);
}
}Test
public class Test {
public static void main(String[] args) {
Student student1 = new Student(1,"小朱");
Student student2 = new Student(2,"大牛");
HashMap<Student,String> hashMap = new HashMap<>();
hashMap.put(student1,"菜鸡");
hashMap.put(student2,"大神");
//这个方法不需要穿Student,为了验证是同一个对象所以改动了一下
updateValue(hashMap,student1);
System.out.println(hashMap.size());
}
public static HashMap<Student,String> updateValue(HashMap<Student,String> hashMap,Student student){
Student student1 = new Student(1,"小朱");
System.out.println(student1.equals(student));
hashMap.put(student1,"新手");
return hashMap;
}
}运行结果
true
3小朱: 代码就是这样,为了验证这个两个对象是相同的,我该遭了updateValue方法.本来只要传HashMap<Student,String>,为了验证我外加了一个Student.
大牛: 哦,确实有意思,你怎么不尝试写个取出元素的方法,例如:
getValue
public static void getValue(HashMap<Student,String> hashMap){
Student student1 = new Student(1,"小朱");
System.out.println(hashMap.get(student1));
}小朱: 我试下.
输出结果
null小朱: 为什么?我重写了equals它们对象都相等啊?为什么取不出来!
大牛: HashMap源码了解一下.
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
/**
* Implements Map.put and related methods.
*
* @param hash hash for key
* @param key the key
* @param value the value to put
* @param onlyIfAbsent if true, don't change existing value
* @param evict if false, the table is in creation mode.
* @return previous value, or null if none
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}大牛: 还有下面这段.
public static void main(String[] args) {
Student student1 = new Student(1,"小朱");
Student student2 = new Student(1,"小朱");
System.out.println(student1.equals(student2));
System.out.println(student1.hashCode());
System.out.println(student2.hashCode());
}小朱: 我跑一下看看.
true
1456208737
288665596小朱: 我明白了,HashMap先判断hashCode值,如果不相同则算这两个对象不相同,如果相同防止哈希冲突,再去判断equals,我这边只重写了equals方法,而hashCode没重写才出现了这个问题.
大牛: 只重写了equals方法,而hashCode不重写,违背了java规范.虽然某些情况下可以,但是用在HashSet,HashMap等等地方就会出现问题.所以不要偷懒哦.
小朱: 学到了.
各位读者根据这个题目,可以探索出一个面试题.
student1.equals(student2) //为true时,
hashMap.put(student1,"菜鸡");
hashMap.put(student2,"大神");
System.out.println(hashMap.size()); //是多少?去考考你们的小伙伴吧.
