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 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.  *  * 
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.)  *  * 
A return value of {@code null} does not necessarily  * 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 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 getNode(int hash, Object key) {     Node[] tab; Node 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)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; }