How to modify the value in Hashmap based on key value in Java

Modify the value of Hashmap according to the key value

If there is no key in the original map, it will be created. If there is a key in the original map, the original value will be overwritten with value. The value of

map.put(key,value);

This implementation adds one to the original value (provided there is this key)

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

The value corresponding to the key can be obtained as follows, if not, the default value can be returned

map.getOrDefault(key,value);

Can the value be obtained correctly after the key of HashMap is changed?

A series of key-value pairs stored in HashMap, where the key is a custom type. After putting it into HashMap, we change the attributes of a certain key externally, and then we use this key to take out elements from HashMap. What will HashMap return at this time?

The answers from several people in our office are inconsistent. Some say it returns null, and some say it can return value normally. But whatever the answer is, there is no solid reason. I thought this question was quite interesting, so I wrote a code test. The result is null. It should be noted that our custom class overrides the hashCode method. I think this result is a bit unexpected, because we know that HashMap stores reference types, and we updated the key outside, which means that the key inside the HashMap is also updated, which means that the hashCode return value of this key will also happen. Variety. At this time, the hashCode of key and HashMap must be the same as the hashCode of the element, and equals will definitely return true, because it is the same object, so why can't it return the correct value?

Test Case

There are 2 cases here, one is the Person class, and the other is the Student class. Let’s verify the above point of view (with conclusion):

• Whether modifying the properties of an object will change its hashcode => Yes

• Whether the value will be affected by modifying the properties when accessing the HashMap=> ; The value is 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 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;
}

Output result

123456
person
123456
person
71154
student
71213
null

Source code

hashCode source code

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 source code

/**
 * 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;
}

In short

You can see that a table is obtained first, and this table is actually an array. Then find the value corresponding to key in the table. The criterion for finding is that the hash is equal to the hash of the passed-in parameter, and meets one of the other two conditions: k = e.key, which means they are the same object, or the key of the equal target of the passed-in key. Our problem lies in the hash(key.hashCode()). You can see that HashMap hashes the hashCode of key when storing elements. The obtained hash will eventually be used as the basis for the storage location of the element. Corresponding to our situation: when storing for the first time, the hash function uses key.hashCode as a parameter to get a value, and then stores the element in a certain location based on this value.

When we fetch the element again, the value of key.hashCode has changed, so the hash function result here has also changed, so when it tries to obtain the storage location of this key, it cannot Get the correct value, resulting in the target element not being found. To return it correctly, it is very simple. Change the hashCode method of the Person class so that its hashCode does not depend on the attribute we want to modify. However, this must not be done in actual development. We always hope that when the attributes of the two objects are different, Different hashCode values ​​can be returned when they are exactly the same.

So the conclusion is that after placing the object in a HashMap, do not modify the attributes of the key unless you override the hashCode method of the entity class which is not subject to attribute restrictions.

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