对象作为HashMap的key

偏执的太偏执、 2021-09-30 12:40 276阅读 0赞

在实际使用中如果遇到对象作为HashMap结构的key，则一定要注意重写equals和hashCode两个方法。以JDK8为例，HashMap在put(K key, V value)方法和containsKey(Object key)方法都会进行当前HashMap中是否已有待放入的key，比如containsKey(Object key)方法：

/**
         * Returns <tt>true</tt> if this map contains a mapping for the
         * specified key.
         *
         * @param   key   The key whose presence in this map is to be tested
         * @return <tt>true</tt> if this map contains a mapping for the specified
         * key.
         */
        public boolean containsKey(Object key) {
            return getNode(hash(key), key) != null;
        }
    
        /**
         * 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;
        }

可以看到，在判断当前Map中是否有要查询的Object key时，先进入混淆函数hash(Object key)计算得出对应的hash值，再根据hash值和equals()函数结果判断是否有当前key。所以，尤其在自定义对象作为key的时候，一定要覆盖equals和hashCode两个方法，equals方法容易实现，hashCode方法就需要开发人员合理地设计了，因为要遵循两个对象equals为true，那么这两个对象中的每个对象调用 `hashCode` 方法都必须生成相同的整数结果。

测试代码：

给学生发学生卡，每个学生对应一个唯一的学号

/**
     * @author lingyingqiaoren
     * @since 2019-08-02
     */
    public class Student {
    
        /**
         * 学号
         */
        private Integer id;
    
        /**
         * 姓名
         */
        private String name;
    
        public Student() {
        }
    
        public Student(Integer id, String name) {
            this.id = id;
            this.name = name;
        }
    
        public Integer getId() {
            return this.id;
        }
    
        @Override
        public String toString(){
            return "Student{id:" + this.id + ";name:" + this.name + "}";
        }
    
        @Override
        public boolean equals(Object o){
            return this.id.equals(((Student) o).getId());
        }
    
        @Override
        public int hashCode(){
            return this.id;
        }
    }

import java.util.ArrayList;
    import java.util.HashMap;
    import java.util.List;
    import java.util.Map;
    
    public class Main {
    
        public static void main(String[] args) {
            List<Student> students = new ArrayList<>();
            Student xiaoMing = new Student(1,"小明");
            //故意把小亮学号设成小明的，作为异常数据，不应该发给他学生卡
            Student xiaoLiang = new Student(1,"小亮");
            students.add(xiaoMing);
            students.add(xiaoLiang);
            Map<Student, SchoolCard> match = new HashMap<>(16);
            students.stream().forEach(
                    student -> {
                        if (!match.containsKey(student)) {
                            match.put(student, new SchoolCard());
                        }
                    }
            );
            System.out.println(match);
        }
    }

程序运行输出结果：

{Student{id:1;name:小明}=SchoolCard@4f3f5b24}

如果注释掉Student类中equals和hashCode两个方法中的任意一个，程序就会输出如下：

{Student{id:1;name:小亮}=SchoolCard@7b23ec81, Student{id:1;name:小明}=SchoolCard@6acbcfc0}