Develop Reference

How does a HashMap use equals() for ArrayList<Integer>? [duplicate]

Is it bad practice to use mutable objects as Hashmap keys? What happens when you try to retrieve a value from a Hashmap using a key that has been modified enough to change its hashcode?
For example, given
class Key
{
int a; //mutable field
int b; //mutable field
public int hashcode()
return foo(a, b);
// setters setA and setB omitted for brevity
}
with code
HashMap<Key, Value> map = new HashMap<Key, Value>();
Key key1 = new Key(0, 0);
map.put(key1, value1); // value1 is an instance of Value
key1.setA(5);
key1.setB(10);
What happens if we now call map.get(key1)? Is this safe or advisable? Or is the behavior dependent on the language?

It has been noted by many well respected developers such as Brian Goetz and Josh Bloch that :
If an object’s hashCode() value can change based on its state, then we
must be careful when using such objects as keys in hash-based
collections to ensure that we don’t allow their state to change when
they are being used as hash keys. All hash-based collections assume
that an object’s hash value does not change while it is in use as a
key in the collection. If a key’s hash code were to change while it
was in a collection, some unpredictable and confusing consequences
could follow. This is usually not a problem in practice — it is not
common practice to use a mutable object like a List as a key in a
HashMap.

This is not safe or advisable. The value mapped to by key1 can never be retrieved. When doing a retrieval, most hash maps will do something like
Object get(Object key) {
int hash = key.hashCode();
//simplified, ignores hash collisions,
Entry entry = getEntry(hash);
if(entry != null && entry.getKey().equals(key)) {
return entry.getValue();
}
return null;
}
In this example, key1.hashcode() now points to the wrong bucket of the hash table, and you will not be able to retrieve value1 with key1.
If you had done something like,
Key key1 = new Key(0, 0);
map.put(key1, value1);
key1.setA(5);
Key key2 = new Key(0, 0);
map.get(key2);
This will also not retrieve value1, as key1 and key2 are no longer equal, so this check
if(entry != null && entry.getKey().equals(key))
will fail.

Hash maps use hash code and equality comparisons to identify a certain key-value pair with a given key. If the has map keeps the key as a reference to the mutable object, it would work in the cases where the same instance is used to retrieve the value. Consider however, the following case:
T keyOne = ...;
T keyTwo = ...;
// At this point keyOne and keyTwo are different instances and
// keyOne.equals(keyTwo) is true.
HashMap myMap = new HashMap();
myMap.push(keyOne, "Hello");
String s1 = (String) myMap.get(keyOne); // s1 is "Hello"
String s2 = (String) myMap.get(keyTwo); // s2 is "Hello"
// because keyOne equals keyTwo
mutate(keyOne);
s1 = myMap.get(keyOne); // returns "Hello"
s2 = myMap.get(keyTwo); // not found
The above is true if the key is stored as a reference. In Java usually this is the case. In .NET for instance, if the key is a value type (always passed by value), the result will be different:
T keyOne = ...;
T keyTwo = ...;
// At this point keyOne and keyTwo are different instances
// and keyOne.equals(keyTwo) is true.
Dictionary myMap = new Dictionary();
myMap.Add(keyOne, "Hello");
String s1 = (String) myMap[keyOne]; // s1 is "Hello"
String s2 = (String) myMap[keyTwo]; // s2 is "Hello"
// because keyOne equals keyTwo
mutate(keyOne);
s1 = myMap[keyOne]; // not found
s2 = myMap[keyTwo]; // returns "Hello"
Other technologies might have other different behaviors. However, almost all of them would come to a situation where the result of using mutable keys is not deterministic, which is very very bad situation in an application - a hard to debug and even harder to understand.

If key’s hash code changes after the key-value pair (Entry) is stored in HashMap, the map will not be able to retrieve the Entry.
Key’s hashcode can change if the key object is mutable. Mutable keys in HahsMap can result in data loss.

This will not work. You are changing the key value, so you are basically throwing it away. Its like creating a real life key and lock, and then changing the key and trying to put it back in the lock.

As others explained, it is dangerous.
A way to avoid that is to have a const field giving explicitly the hash in your mutable objects (so you would hash on their "identity", not their "state"). You might even initialize that hash field more or less randomly.
Another trick would be to use the address, e.g. (intptr_t) reinterpret_cast<void*>(this) as a basis for hash.
In all cases, you have to give up hashing the changing state of the object.

There are two very different issues that can arise with a mutable key depending on your expectation of behavior.
First Problem: (probably most trivial--but hell it gave me problems that I didn't think about!)
You are attempting to place key-value pairs into a map by updating and modifying the same key object. You might do something like Map<Integer, String> and simply say:
int key = 0;
loop {
map.put(key++, newString);
}
I'm reusing the "object" key to create a map. This works fine in Java because of autoboxing where each new value of key gets autoboxed to a new Integer object. What would not work is if I created my own (mutable) Integer object:
MyInteger {
int value;
plusOne(){
value++;
}
}
Then tried the same approach:
MyInteger key = new MyInteger(0);
loop{
map.put(key.plusOne(), newString)
}
My expectation is that, for instance, I map 0 -> "a" and 1 -> "b". In the first example, if I change int key = 0, the map will (correctly) give me "a". For simplicity let's assume MyInteger just always returns the same hashCode() (if you can somehow manage to create unique hashCode values for all possible states of an object, this will not be an issue, and you deserve an award). In this case, I call 0 -> "a", so now the map holds my key and maps it to "a", I then modify key = 1 and try to put 1 -> "b". We have a problem! The hashCode() is the same, and the only key in the HashMap is my MyInteger key object which has just been modified to be equal to 1, so It overwrites that key's value so that now, instead of a map with 0 -> "a" and 1 -> "b", I have 1 -> "b" only! Even worse, if I change back to key = 0, the hashCode points to 1 -> "b", but since the HashMap's only key is my key object, it satisfied the equality check and returns "b", not "a" as expected.
If, like me, you fall prey to this type of issue, it's incredibly difficult to diagnose. Why? Because if you have a decent hashCode() function it will generate (mostly) unique values. The hash value will largely take care of the inequality problem when structuring the map but if you have enough values, eventually you'll get a collision on the hash value and then you get unexpected and largely inexplicable results. The resultant behavior is that it works for small runs but fails for larger ones.
Advice:
To find this type of issue, modify the hashCode() method, even trivially (i.e. = 0--obviously when doing this, keep in mind that the hash values should be the same for two equal objects*), and see if you get the same results--because you should and if you don't, there's likely a semantic error with your implementation that's using a hash table.
*There should be no danger (if there is--you have a semantic problem) in always returning 0 from a hashCode() (although it would defeat the purpose of a Hash Table). But that's sort of the point: the hashCode is a "quick and easy" equality measure that's not exact. So two very different objects could have the same hashCode() yet not be equal. On the other hand, two equal objects must always have the same hashCode() value.
p.s. In Java, from my understanding, if you do such a terrible thing (as have many hashCode() collisions), it will start using a red-black-tree as opposed to ArrayList. So when you expect O(1) lookup, you'll get O(log(n))--which is better than the ArrayList which would give O(n).
Second Problem:
This is the one that most others seem to be focusing on, so I'll try to be brief. In this use case, I try to map a key-value pair and then I do some work on the key and then want to come back and get my value.
Expectation: key -> value is mapped, I then modify key and try to get(key). I expect that will give me value.
It seems kind of obvious to me that this wouldn't work but I'm not above having tried to use things like Collections as a key before (and quite quickly realizing it doesn't work). It doesn't work because it's quite likely that the hash value of key has changed so you won't even be looking in the correct bucket.
This is why it's very inadvisable to use collections as keys. I would assume, if you were doing this, you're trying to establish a many-to-one relationship. So I have a class (as in teaching) and I want two groups to do two different projects. What I want is that given a group, what is their project? Simple, I divide the class in two, and I have group1 -> project1 and group2 -> project2. But wait! A new student arrives so I place them in group1. The problem is that group1 has now been modified and likely its hash value has changed, therefore trying to do get(group1) is likely to fail because it will look in a wrong or non-existent bucket of the HashMap.
The obvious solution to the above is to chain things--instead of using the groups as keys, give them labels (that don't change) that point to the group and therefore the project: g1 -> group1 and g1 -> project1, etc.
p.s.
Please make sure to define a hashCode() and equals(...) method for any object you expect to use as a key (eclipse and, I'm assuming, most IDE's can do this for you).
Code Example:
Here is a class which exhibits the two different "problem" behaviors. In this case, I attempt to map 0 -> "a", 1 -> "b", and 2 -> "c" (in each case). In the first problem, I do that by modifying the same object, in the second problem, I use unique objects, and in the second problem "fixed" I clone those unique objects. After that I take one of the "unique" keys (k0) and modify it to attempt to access the map. I expect this will give me a, b, c and null when the key is 3.
However, what happens is the following:
map.get(0) map1: 0 -> null, map2: 0 -> a, map3: 0 -> a
map.get(1) map1: 1 -> null, map2: 1 -> b, map3: 1 -> b
map.get(2) map1: 2 -> c, map2: 2 -> a, map3: 2 -> c
map.get(3) map1: 3 -> null, map2: 3 -> null, map3: 3 -> null
The first map ("first problem") fails because it only holds a single key, which was last updated and placed to equal 2, hence why it correctly returns "c" when k0 = 2 but returns null for the other two (the single key doesn't equal 0 or 1). The second map fails twice: the most obvious is that it returns "b" when I asked for k0 (because it's been modified--that's the "second problem" which seems kind of obvious when you do something like this). It fails a second time when it returns "a" after modifying k0 = 2 (which I would expect to be "c"). This is more due to the "first problem": there's a hash code collision and the tiebreaker is an equality check--but the map holds k0, which it (apparently for me--could theoretically be different for someone else) checked first and thus returned the first value, "a" even though had it kept checking, "c" would have also been a match. Finally, the 3rd map works perfectly because I'm enforcing that the map holds unique keys no matter what else I do (by cloning the object during insertion).
I want to make clear that I agree, cloning is not a solution! I simply added that as an example of why a map needs unique keys and how enforcing unique keys "fixes" the issue.
public class HashMapProblems {
private int value = 0;
public HashMapProblems() {
this(0);
}
public HashMapProblems(final int value) {
super();
this.value = value;
}
public void setValue(final int i) {
this.value = i;
}
#Override
public int hashCode() {
return value % 2;
}
#Override
public boolean equals(final Object o) {
return o instanceof HashMapProblems
&& value == ((HashMapProblems) o).value;
}
#Override
public Object clone() {
return new HashMapProblems(value);
}
public void reset() {
this.value = 0;
}
public static void main(String[] args) {
final HashMapProblems k0 = new HashMapProblems(0);
final HashMapProblems k1 = new HashMapProblems(1);
final HashMapProblems k2 = new HashMapProblems(2);
final HashMapProblems k = new HashMapProblems();
final HashMap<HashMapProblems, String> map1 = firstProblem(k);
final HashMap<HashMapProblems, String> map2 = secondProblem(k0, k1, k2);
final HashMap<HashMapProblems, String> map3 = secondProblemFixed(k0, k1, k2);
for (int i = 0; i < 4; ++i) {
k0.setValue(i);
System.out.printf(
"map.get(%d) map1: %d -> %s, map2: %d -> %s, map3: %d -> %s",
i, i, map1.get(k0), i, map2.get(k0), i, map3.get(k0));
System.out.println();
}
}
private static HashMap<HashMapProblems, String> firstProblem(
final HashMapProblems start) {
start.reset();
final HashMap<HashMapProblems, String> map = new HashMap<>();
map.put(start, "a");
start.setValue(1);
map.put(start, "b");
start.setValue(2);
map.put(start, "c");
return map;
}
private static HashMap<HashMapProblems, String> secondProblem(
final HashMapProblems... keys) {
final HashMap<HashMapProblems, String> map = new HashMap<>();
IntStream.range(0, keys.length).forEach(
index -> map.put(keys[index], "" + (char) ('a' + index)));
return map;
}
private static HashMap<HashMapProblems, String> secondProblemFixed(
final HashMapProblems... keys) {
final HashMap<HashMapProblems, String> map = new HashMap<>();
IntStream.range(0, keys.length)
.forEach(index -> map.put((HashMapProblems) keys[index].clone(),
"" + (char) ('a' + index)));
return map;
}
}
Some Notes:
In the above it should be noted that map1 only holds two values because of the way I set up the hashCode() function to split odds and evens. k = 0 and k = 2 therefore have the same hashCode of 0. So when I modify k = 2 and attempt to k -> "c" the mapping k -> "a" gets overwritten--k -> "b" is still there because it exists in a different bucket.
Also there are a lot of different ways to examine the maps in the above code and I would encourage people that are curious to do things like print out the values of the map and then the key to value mappings (you may be surprised by the results you get). Do things like play with changing the different "unique" keys (i.e. k0, k1, and k2), try changing the single key k. You could also see how even the secondProblemFixed isn't actually fixed because you could also gain access to the keys (for instance via Map::keySet) and modify them.

I won't repeat what others have said. Yes, it's inadvisable. But in my opinion, it's not overly obvious where the documentation states this.
You can find it on the JavaDoc for the Map interface:
Note: great care must be exercised if mutable objects are used as map
keys. The behavior of a map is not specified if the value of an object
is changed in a manner that affects equals comparisons while the
object is a key in the map

Behaviour of a Map is not specified if value of an object is changed in a manner that affects equals comparision while object(Mutable) is a key. Even for Set also using mutable object as key is not a good idea.
Lets see a example here :
public class MapKeyShouldntBeMutable {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Map<Employee,Integer> map=new HashMap<Employee,Integer>();
Employee e=new Employee();
Employee e1=new Employee();
Employee e2=new Employee();
Employee e3=new Employee();
Employee e4=new Employee();
e.setName("one");
e1.setName("one");
e2.setName("three");
e3.setName("four");
e4.setName("five");
map.put(e, 24);
map.put(e1, 25);
map.put(e2, 26);
map.put(e3, 27);
map.put(e4, 28);
e2.setName("one");
System.out.println(" is e equals e1 "+e.equals(e1));
System.out.println(map);
for(Employee s:map.keySet())
{
System.out.println("key : "+s.getName()+":value : "+map.get(s));
}
}
}
class Employee{
String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#Override
public boolean equals(Object o){
Employee e=(Employee)o;
if(this.name.equalsIgnoreCase(e.getName()))
{
return true;
}
return false;
}
public int hashCode() {
int sum=0;
if(this.name!=null)
{
for(int i=0;i<this.name.toCharArray().length;i++)
{
sum=sum+(int)this.name.toCharArray()[i];
}
/*System.out.println("name :"+this.name+" code : "+sum);*/
}
return sum;
}
}
Here we are trying to add mutable object "Employee" to a map. It will work good if all keys added are distinct.Here I have overridden equals and hashcode for employee class.
See first I have added "e" and then "e1". For both of them equals() will be true and hashcode will be same. So map sees as if the same key is getting added so it should replace the old value with e1's value. Then we have added e2,e3,e4 we are fine as of now.
But when we are changing the value of an already added key i.e "e2" as one ,it becomes a key similar to one added earlier. Now the map will behave wired. Ideally e2 should replace the existing same key i.e e1.But now map takes this as well. And you will get this in o/p :
is e equals e1 true
{Employee#1aa=28, Employee#1bc=27, Employee#142=25, Employee#142=26}
key : five:value : 28
key : four:value : 27
key : one:value : 25
key : one:value : 25
See here both keys having one showing same value also. So its unexpected.Now run the same programme again by changing e2.setName("diffnt"); which is e2.setName("one"); here ...Now the o/p will be this :
is e equals e1 true
{Employee#1aa=28, Employee#1bc=27, Employee#142=25, Employee#27b=26}
key : five:value : 28
key : four:value : 27
key : one:value : 25
key : diffnt:value : null
So by adding changing the mutable key in a map is not encouraged.

To make the answer compact:
The root cause is that HashMap calculates an internal hash of the user's key object hashcode only once and stores it inside for own needs.
All other operations for data navigation inside the map are doing by this pre-calculated internal hash.
So if you change the hashcode of the key object (mutate) it will be still stored nicely inside the map with the changed key object's hashcode (you could even observe it via HashMap.keySet() and see the altered hashcode).
But HashMap internal hash will not be recalculated of course and it will be the old stored one and the map won't be able to locate your data by the provided mutated key object new hashcode. (e.g. by HashMap.get() or HashMap.containsKey()).
Your key-value pairs will be still inside the map but to get it back you will need that old hash code value that was given when you put your data into the map.
Notice that you also will be unable to get data back by the mutated key object taken right from the HashMap.keySet().

Can we use object as a key in hashmap in Java?

How to use an object as a key in a hashmap. If you use an object as key do you need to override equals and hashcode methods for that object?

A simple thumb rule is to use immutable objects as keys in a HashMap.
because:
If it were mutable, then the hashcode() value or equals() condition might change, and you would never be able to retrieve the key from your HashMap.
More precisely, class fields that are used to compute equals() and hashcode() should be immutable!
Now, suppose you create your own class:
To compare two objects of your class you will have to override equals()
To use it as a key in any Hash based Data structure you will have to override hashcode() (again, keeping immutability in mind)
Remember that if two objects are equal(), then their hashcode() should be equal as well!

hashCode() -HashMap provides put(key, value) for storing and get(key) for retrieving values from a HashMap. When using put(key, value) to store a key-value-pair, HashMap calls hashcode() on the key object to calculate a hash that is used to find a bucket where the Entry object is stored. When get() is used to retrieve a value, again, the key object is used to calculate a hash which is used then to find a bucket where that particular key is stored.
equals() - equals() is used to compare objects for equality. In the case of HashMap, the key object is used for comparison, also using equals(). HashMap knows how to handle hashing collisions (more than one key having the same hash value, thus assigned to the same bucket). In that case objects are stored in a linked list (refer to the figure for more clarity).
hashCode() helps in finding the bucket where that key is stored, equals() helps in finding the right key as there may be more than one key-value pair stored in a single bucket.

You can use any object in a HashMap as long as it has properly defined hashCode and equals methods - those are absolutely crucial because the hashing mechanism depends on them.

Answer to your question is yes, objects of custom classes can be used as a key in a HashMap. But in order to retrieve the value object back from the map without failure, there are certain guidelines that need to be followed.
1)Custom class should follow the contract between hashCode() and equals().
The contract states that:
If two objects are equal according to the equals(Object) method, then calling
the hashCode method on each of the two objects must produce the same integer result.
This can be done by implementing hashcode() and equals() in your custom class.
2) Make custom class immutable.
Hint: use final, remove setters, use deep copy to set fields

package com.java.demo.map;
import java.util.HashMap;
public class TestMutableKey
{
public static void main(String[] args)
{
//Create a HashMap with mutable key
HashMap<Account, String> map = new HashMap<Account, String>();
//Create key 1
Account a1 = new Account(1);
a1.setHolderName("A_ONE");
//Create key 2
Account a2 = new Account(2);
a2.setHolderName("A_TWO");
//Put mutable key and value in map
map.put(a1, a1.getHolderName());
map.put(a2, a2.getHolderName());
//Change the keys state so hash map should be calculated again
a1.setHolderName("Defaulter");
a2.setHolderName("Bankrupt");
//Success !! We are able to get back the values
System.out.println(map.get(a1)); //Prints A_ONE
System.out.println(map.get(a2)); //Prints A_TWO
//Try with newly created key with same account number
Account a3 = new Account(1);
a3.setHolderName("A_THREE");
//Success !! We are still able to get back the value for account number 1
System.out.println(map.get(a3)); //Prints A_ONE
}
}

Yes, you should override equals and hashcode, for the proper functioning of the code otherwise you won't be able to get the value of the key which you have inserted in the map.
e.g
map.put(new Object() , "value") ;
when you want to get that value ,
map.get(new Object()) ; // This will always return null
Because with new Object() - new hashcode will be generated and it will not point to the expected bucket number on which value is saved, and if eventually bucket number comes to be same - it won't be able to match hashcode and even equals so it always return NULL .

Yes, we can use any object as key in a Map in java but we need to override the equals() and hashCode() methods of that object class. Please refer an example below, in which I am storing an object of Pair class as key in a hashMap with value type as string in map. I have overriden the hashCode() and equals() methods of Pair class. So, that different objects of Pair class with same values of Pair(x,y) will be treated as one object only.
import java.util.*;
import java.util.Map.Entry;
class App { // Case-sensitive
private class Pair {
private int x, y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + x;
result = prime * result + y;
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Pair other = (Pair) obj;
if (x != other.x)
return false;
if (y != other.y)
return false;
return true;
}
}
public static void main(String[] args) {
App obj = new App();
obj.show();
}
private void show() {
Map<Pair, String> map = new HashMap<>();
Pair obj1 = new Pair(10, 20);
Pair obj2 = new Pair(40, 50);
Pair obj3 = new Pair(10, 20);
// We can see that obj1 and obj3 have same values. So, we want to store these
// objects
// as one .To achieve
// that,
// we have overridden equals() and hashCode() methods of Pair class.
map.put(obj1, "First");
map.put(obj2, "Second");
map.put(obj3, "Third");
System.out.printf("Size of Map is :%d \n", map.size());
for (Entry<App.Pair, String> p : map.entrySet()) {
Pair pair = p.getKey();
System.out.printf("Map key-value pair is (%d,%d)->%s \n", pair.x, pair.y, p.getValue());
}
// output -
// Size of Map is :2
// Map key-value pair is (10,20)->Third
// Map key-value pair is (40,50)->Second
}
}

How does overriding equals and hashCode impacts storing data in a Map in Java with a bad hashing function?

I have a class, Employee, let's say, and my hashCode function for this class is really bad (let's say it always return a constant). My code looks like the following.
public class Employee {
private String name;
public Employee(String name) {
this.name = name;
}
#Override
public int hashCode() { return 1; }
#Override
public boolean equals(Object object) {
if(null == object || !(object instanceof Employee)) {
return false;
}
Employee other = (Employee)object;
return this.name.equals(other.name);
}
}
Let's say I want to use Employee as the key in a Map, and so I can do something like the following.
public static void main(String[] args) {
Map<Employee, Long> map = new HashMap<>();
for(int i=0; i < 1000; i++) {
map.put(new Employee("john"+i, 1L));
}
System.out.println(map.size());
}
How come when I run this code, I always get 1,000 as the size?
Using Employee as a key seems to be "good" in the following sense.
It is immutable
Two employees that are equals always generate the same hash code
What I expected was that since the output of hashCode is always 1, then map.size() should always be 1. But it is not. Why? If I have a Map<Integer,Integer>, and I do map.put(1, 1) followed by map.put(1, 2), I would only expect the size to be 1.
The equals method must somehow be coming into play here, but I'm not sure how.
Any pointers are appreciated.

Your loop
for(int i=0; i < 1000; i++) {
map.put(new Employee("john"+System.currentTimeMillis(), 1L));
}
executes within a couple of milliseconds, so System.currentTimeMillis() will be returning the same value for the vast majority of the iterations of your loop. So, several hundred of your johns will have the exact same name + number.
Then, we have java's retarded Map which does not have an add() method, (which one would reasonably expect to throw an exception if the item already exists,) but instead it only has a put() method which will either add or replace items, without failing. So, most of your johns get overwritten by subsequent johns, without any increase in the map size, and without any exception being thrown to give you a hint about what you are doing wrong.
Furthermore, you seem to be a bit confused as to exactly what the effect of a bad hashCode() function is on a map. A bad hashCode() simply results in collisions. Collisions in a hashmap do not cause items to be lost; they only cause the internal structure of the map to not be very efficient. Essentially, a constant hashCode() will result in a degenerate map which internally looks like a linked list. It will be inefficient both for insertions and for deletions, but no items will be lost due to that.
Items will be lost due to a bad equals() method, or due to overwriting them with newer items. (Which is the case in your code.)

Mike's answer is right about what is causing this. But the real reason that it's happening is this:
In the put method of HashMap it first checks the hashcode for each entry. If the hash code is equal to the hashcode of your new key then it checks for .equals(). If equals() returns true it just replaces the existing object with the new one otherwise adds a new key value pair. That's where it gets busted. Because somethings your equals() function will return true because of the currentMilliSeconds and sometimes it won't hence different sizes every time.
Just pay attention to the equals in the code below (java HashMap).
public V put(K key, V value) {
if (key == null)
return putForNullKey(value);
int hash = hash(key.hashCode());
int i = indexFor(hash, table.length);
for (Entry<K,V> e = table[i]; e != null; e = e.next) {
Object k;
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
V oldValue = e.value;
e.value = value;
e.recordAccess(this);
return oldValue;
}
}
modCount++;
addEntry(hash, key, value, i);
return null;
}

If your hashcode is the same for every entry then your time complexity will be O(n) because the hashcode creates buckets to store your elements. If you only create a single bucket then you have to traverse the entire bucket to get your element.
If however, your hashcode is unique for every element then you will have a unique bucket and will only have to traverse a single element.
Bucket lookups (Hash) are O(1) so the better the hashcode the better the time complexity.

I think you have a misconecption what HashBuckets in a HashMap are for.
When you put two Objectswhich are not equal but have the same hashCode in a HashMap, both elements will be present in the Hashmap in the same HashBucket. An element is only overwritten when an element exists in the HashMapwhich has the same hashCode and is equals to an existing element.
The HashBuckets make the HashMap fast at lookup, because when searching for an element, only elements in the HahsBucket corresponding to the hashCode need to be considered. This is why it is generally a bad idea to wirte a HashFunction which is constant.

Your hashcode has to comply with certain requirements e.g equal objects should return equal hashcode.
But your implementation is not solid then it will give performance problem, if many of your objects have the same hashcode your look up some simply become O(N) instead of O(1). In your case it simply putting all items in a List. So the size is 1000.

Duplicate element search and replacement logic in HashMap

This question was asked to me in a job interview and I still don't know answer so ask here. Lets say hashCode() of key object returns a fixed integer so HashMap would look like a LinkedList.
How would a duplicate element be found and replaced by new value in map?
e.g. if following 1001 puts are performed in order listed below,
put(1000,1000), put(1,1), put( 2, 2), put ( 3,3 ) ....put(999,999), put(1000,1000 )
Would map be traversed all the way to end and then new one be inserted at head when last put(1000,1000) is performed?
OR
Map has some other way to locate and replace duplicate keys?

First case is correct.
In your case when hashCode() is returning same hash value for all the keys. In the java HashMap, Key and Value both are stored in the bucket as Map.Entry object. When perform the second or further put() operations into the map, it will traverse all the element to check whether Key is already present in the Map. If Key is not found then new Key and Value pair will be added into the linked list. If Key is found in the list then it update the Value for the pair.
Details explanation about java HashMap working: How HashMap works in Java
Take this sample code and run in the debug mode and observe how the new Key and Value pair are inserted into the Map.
In the class you will need to hashCode() (we want to control how the hash codes are generated for Node), toString() (just to output the Node value in SOUT) and equals() (defines the equality of the keys based on the value of Node member variable Integer, for updating the values.) methods for getting it working.
public class HashMapTest {
static class Node {
Integer n;
public Node(int n) {
this.n = n;
}
#Override
public int hashCode() {
return n%3;
}
#Override
public boolean equals(Object object) {
Node node = (Node)object;
return this.n.equals(node.n);
}
#Override
public String toString() {
return n.toString();
}
}
public static void main(String[] args) {
Map<Node, String> map = new HashMap<>();
for (int i = 0; i<6; i++) {
map.put(new Node(i), ""+i); // <-- Debug Point
}
map.put(new Node(0), "xxx");
} // <-- Debug Point
}
First 3 entries in the map: (hash code is n%3)
Three more values: (hash code is n%3)
Now don't confused about the ordering of the node, I have executed them on java 1.8 and HashMap uses TreeNode, an implementation of Red-Black tree as per the code documentation. This can be different in different versions of the java.
Now lets update the Value of Key 0:

When the hash code is the same, the hash map compares objects using the equals method.
For example, let's say you put a bunch of elements in the hash map:
put(1000,1000), put(1,1), put( 2, 2), put ( 3,3 ) ....put(999,999)
And then you do this:
put(1000,1000 )
1000 is already in the map, the hash code is the same, it is also the same in terms of the equals method, so the element will be replaced, no need to iterate further.
Now if you do:
put(1234, 1234)
1234 is not yet in the map. All the elements are in a linked list, due to the fixed hash code. The hash map will iterate over the elements, comparing them using equals. It will be false for all elements, the end of the list will be reached, and the entry will be appended.

JDK implementations changes over time !
In JDK8, if the hashCode() is a constant value, the implementation creates a tree not a linked list in order to protect against DDOS attack 1.

Why are keys immutable in Java?

Apologies for the fairly naive question, but I believe my own answer to be naive. I think keys (in HashTables) are immutable because we wouldn't want to somehow accidentally alter a key and therefore mess with the sorting of the HashTable. Is this a correct explanation? If so, how can it be more correct?

During the HashTable.put the key is hashed and it an its value are stored in one of a number of buckets (which are lists of key value pairs) based on the hash, for example something like:
bucket[key.hashcode() % numberOfBuckets].add(key, value)
If the key's hashcode changes after insertion it could then be in the wrong bucket and you would then not be able to find it and the hashtable would incorrectly return null on any get for that key.
Aside: Understanding the inner workings of a hashtable helps you understand the importance of a good quality hashcode function for your keys. As a poor hashcode function could result in a poor distribution of keys in buckets. And as buckets are just lists, this results in a lot of linear searches which greatly reduces the effectiveness of the hashtable. e.g. this terrible hashcode function puts everything in one bucket, so it's effectively just one list.
public int hashcode { return 42; /*terrible hashcode example, don't use!*/ }
This is also one reason why prime numbers appear in good hashcode functions, e.g.:
public int hashcode {
int hash = field1.hashcode();
hash = hash*31 + field2.hashcode(); //note the prime 31
hash = hash*31 + field3.hashcode();
return hash;
}

The general idea is correct, but not its details.
Keys in a HashTable need not be immutable, it's the result of a call to their hashCode() (and equals) method that needs to stay immutable and consistent (for the hash table to behave predictably, that is).
From a high-level point of view, it's because of the way hash tables work : when a (key, value) pair is inserted, the key's hashCode is used internally to figure out a "bucket" where the value will be put. And when the value is retrieved by key, the hashCode is computed once more, to find the bucket back.
Now if at any point in time between the insertion and the retreival, the result of calling hashCode changes, the "lookup bucket" will be different than the "insertion" bucket, and things will not behave predictably.
To sum up, given a Key object that looks like this (two internal Strings compose the objet, but only one, partOfHashCode is taken into account in hashCode / equals) :
public static class Key {
private String partOfHashCode;
private String notPartOfHashCode;
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((partOfHashCode == null) ? 0 : partOfHashCode.hashCode());
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Key other = (Key) obj;
if (partOfHashCode == null) {
if (other.partOfHashCode != null)
return false;
} else if (!partOfHashCode.equals(other.partOfHashCode))
return false;
return true;
}
}
The it is fine to use it this way :
public static void main(String[] args) {
Map<Key, String> myMap = new HashMap<>();
Key key = new Key();
key.partOfHashCode = "myHash";
myMap.put(key, "value");
key.notPartOfHashCode = "mutation of the key, but not of its hash/equals definition";
System.out.println(myMap.get(key));
}
(This logs the "value" object in the console).
But it is NOT fine to use it this way
public static void main(String[] args) {
Map<Key, String> myMap = new HashMap<>();
Key key = new Key();
key.partOfHashCode = "myHash";
myMap.put(key, "value");
key.partOfHashCode = "mutation of the hashCode of the key";
System.out.println(myMap.get(key));
}
(This last example could log "null" in the console).
For more on this subject, you should read also on hashCode / equals consistency.

There is no inherent guarantee in Java that HashTable-keys are immutable. It is not even guaranteed that their hashcode remains the same. But if you add keys that have a mutable hashCode you are in trouble. Assume you are inserting a key with a hashCode of 1. Then it is inserted in a hash-bucket corresponding to 1. Then alter the object to have a hashCode of 2 and call hashMap.get(key). While the object is still in the hashTable the system will look in the bucket corresponding to 2, but won't find it there. You won't even be able to remove the entry since it won't be found.
tl;dr For your application to work properly HashTable-keys need to have immutable hashcodes, but you have to take care of that fact for yourself.