Java如何根据key值修改Hashmap中的value值

map.put(key,value);

map.put(key,map.get(key)+1);

map.getOrDefault(key,value);

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;
}

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;
}

/**
 * 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;
}