Java怎么根据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; }