What happens when a duplicate key is put into a HashMap? – Dev

The best answers to the question “What happens when a duplicate key is put into a HashMap?” in the category Dev.

QUESTION:

If I pass the same key multiple times to HashMap’s put method, what happens to the original value? And what if even the value repeats? I didn’t find any documentation on this.

Case 1: Overwritten values for a key

Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","not one");
mymap.put("1","surely not one");
System.out.println(mymap.get("1"));

We get surely not one.

Case 2: Duplicate value

Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","not one");
mymap.put("1","surely not one");
// The following line was added:
mymap.put("1","one");
System.out.println(mymap.get("1"));

We get one.

But what happens to the other values? I was teaching basics to a student and I was asked this. Is the Map like a bucket where the last value is referenced (but in memory)?

ANSWER:

You may find your answer in the javadoc of Map#put(K, V) (which actually returns something):

public V put(K key,
             V value)

Associates the specified value with the specified key in this map
(optional operation). If the map
previously contained a mapping for
this key, the old value is replaced by
the specified value. (A map m is said
to contain a mapping for a key k if
and only if m.containsKey(k) would
return true.)

Parameters:
key – key with which the specified value is to be associated.
value – value to be associated with the specified key.

Returns:
previous value associated with specified key, or null if there was no
mapping for key. (A null return can also indicate that the map previously associated null with the specified key, if the implementation supports null values.)

So if you don’t assign the returned value when calling mymap.put("1", "a string"), it just becomes unreferenced and thus eligible for garbage collection.

ANSWER:

By definition, the put command replaces the previous value associated with the given key in the map (conceptually like an array indexing operation for primitive types).

The map simply drops its reference to the value. If nothing else holds a reference to the object, that object becomes eligible for garbage collection. Additionally, Java returns any previous value associated with the given key (or null if none present), so you can determine what was there and maintain a reference if necessary.

More information here: HashMap Doc

ANSWER:

It replaces the existing value in the map for the respective key. And if no key exists with the same name then it creates a key with the value provided.
eg:

Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","two");

OUTPUT
key = “1”, value = “two”

So, the previous value gets overwritten.

ANSWER:

it’s Key/Value feature and you could not to have duplicate key for several values because when you want to get the actual value which one of values is belong to entered key
in your example when you want to get value of “1” which one is it ?!
that’s reasons to have unique key for every value but you could to have a trick by java standard lib :

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;

public class DuplicateMap<K, V> {

    private Map<K, ArrayList<V>> m = new HashMap<>();

    public void put(K k, V v) {
        if (m.containsKey(k)) {
            m.get(k).add(v);
        } else {
            ArrayList<V> arr = new ArrayList<>();
            arr.add(v);
            m.put(k, arr);
        }
    }

     public ArrayList<V> get(K k) {
        return m.get(k);
    }

    public V get(K k, int index) {
        return m.get(k).size()-1 < index ? null : m.get(k).get(index);
    }
}

and you could to use it in this way:

    public static void main(String[] args) {
    DuplicateMap<String,String> dm=new DuplicateMap<>();
    dm.put("1", "one");
    dm.put("1", "not one");
    dm.put("1", "surely not one");
    System.out.println(dm.get("1"));
    System.out.println(dm.get("1",1));
    System.out.println(dm.get("1", 5));
}

and result of prints are :

[one, not one, surely not one]
not one
null