Java - HashMap 的 key 更改后能否正确获取 value？

在HashMap 中存放的一系列键值对，其中键为某个我们自定义的类型。放入 HashMap 后，我们在外部把某一个 key 的属性进行更改，然后我们再用这个 key 从 HashMap 里取出元素，这时候 HashMap 会返回什么？

我们办公室几个人答案都不一致，有的说返回null，有的说能正常返回value。但不论答案是什么都没有确凿的理由。我觉得这个问题挺有意思的，就写了代码测试。结果是返回null。需要说明的是我们自定义的类重写了 hashCode 方法。我想这个结果还是有点意外的，因为我们知道 HashMap 存放的是引用类型，我们在外面把 key 更新了，那也就是说 HashMap 里面的 key 也更新了，也就是这个 key 的 hashCode 返回值也会发生变化。这个时候 key 的 hashCode 和 HashMap 对于元素的 hashCode 肯定一样，equals也肯定返回true，因为本来就是同一个对象，那为什么不能返回正确的值呢？

测试案例

这里有 2 个案例，一个是 Person 类，还有一个是 Student 类，我们来验证下以上的观点（附带结论）：

1. 修改了对象属性是否会改变它的 hashcode => 是的
2. 在 HashMap 里存取的时候是否会受到修改属性影响取值 => 取值为 null

package tech.luxsun.interview.luxinterviewstarter.collection;

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
import java.util.HashMap;

/**
 * @author Lux Sun
 * @date 2021/4/22
 */
public class MapDemo0 {

    public static void main(String[] args) {
        HashMap<Object, Object> map = new HashMap<>();

        // Person Case
        Person p = new Person("Bob", 12);
        map.put(p, "person");
        System.out.println(p.hashCode());
        System.out.println(map.get(p));

        p.setAge(13);
        System.out.println(p.hashCode());
        System.out.println(map.get(p));

        // Student Case
        Student stu = new Student("Bob", 12);
        map.put(stu, "student");
        System.out.println(stu.hashCode());
        System.out.println(map.get(stu));

        stu.setAge(13);
        System.out.println(stu.hashCode());
        System.out.println(map.get(stu));
    }
}

@Data
@AllArgsConstructor
@NoArgsConstructor
class Person {
    private String name;
    private Integer age;

    public int hashCode() {
        return 123456;
    }
}

@Data
@AllArgsConstructor
@NoArgsConstructor
class Student {
    private String name;
    private Integer age;
}      

输出结果

123456
person
123456
person
71154
student
71213
null      

源码

• hashCode 源码

public int hashCode() {
    int PRIME = true;
    int result = 1;
    Object $age = this.getAge();
    int result = result * 59 + ($age == null ? 43 : $age.hashCode());
    Object $name = this.getName();
    result = result * 59 + ($name == null ? 43 : $name.hashCode());
    return result;
}      

• map.get 源码

/**
 * Returns the value to which the specified key is mapped,
 * or {@code null} if this map contains no mapping for the key.
 *
 * <p>More formally, if this map contains a mapping from a key
 * {@code k} to a value {@code v} such that {@code (key==null ? k==null :
 * key.equals(k))}, then this method returns {@code v}; otherwise
 * it returns {@code null}.  (There can be at most one such mapping.)
 *
 * <p>A return value of {@code null} does not <i>necessarily</i>
 * indicate that the map contains no mapping for the key; it's also
 * possible that the map explicitly maps the key to {@code null}.
 * The {@link #containsKey containsKey} operation may be used to
 * distinguish these two cases.
 *
 * @see #put(Object, Object)
 */
public V get(Object key) {
    Node<K,V> e;
    return (e = getNode(hash(key), key)) == null ? null : e.value;
}


/**
 * Computes key.hashCode() and spreads (XORs) higher bits of hash
 * to lower.  Because the table uses power-of-two masking, sets of
 * hashes that vary only in bits above the current mask will
 * always collide. (Among known examples are sets of Float keys
 * holding consecutive whole numbers in small tables.)  So we
 * apply a transform that spreads the impact of higher bits
 * downward. There is a tradeoff between speed, utility, and
 * quality of bit-spreading. Because many common sets of hashes
 * are already reasonably distributed (so don't benefit from
 * spreading), and because we use trees to handle large sets of
 * collisions in bins, we just XOR some shifted bits in the
 * cheapest possible way to reduce systematic lossage, as well as
 * to incorporate impact of the highest bits that would otherwise
 * never be used in index calculations because of table bounds.
 */
static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

/**
 * Implements Map.get and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @return the node, or null if none
 */
final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (first = tab[(n - 1) & hash]) != null) {
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        if ((e = first.next) != null) {
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}      

总结

所以结论就是当把对象放到 HashMap 后，不要去修改 key 的属性，除非你重写了该实体类的 hashCode 方法不受属性限制。