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Duplicate elements being added in a set after I change the value after insertion

I have a set with two distinct objects added to it. After insertion, I change one of the objects in such a way that both the objects are equal (as verified by the overridden equals method in object class). At this point in time I have two duplicate elements in a set. Now I try to add these two duplicate objects in a new set and I am still able to add them even though the equals method returns true for them. Below is the code for the same. Can someone please tell me what exactly am I missing?
public class BasicSetImpl{
public int num; String entry;
public BasicSetImpl(int num, String entry){
this.num = num;
this.entry = entry;
}
#Override
public int hashCode() {
return Objects.hash(entry, num);
}
#Override
public boolean equals(Object obj) {
BasicSetImpl newObj = (BasicSetImpl)obj;
if (this.num == newObj.num)
return true;
else
return false;
}
public static void main(String[] args){
Set<BasicSetImpl> set = new HashSet<>();
BasicSetImpl k1 = new BasicSetImpl(1, "One");
BasicSetImpl k2 = new BasicSetImpl(2, "Two");
set.add(k1);
set.add(k2);
k2.num = 1;
System.out.println(k1.equals(k2)); //This line returns True
Set<BasicSetImpl> newSet = new HashSet<>();
newSet.add(k1);
newSet.add(k2);
//Set.size here is two

The hash based collection, in this case the HashSet uses the Object hashCode method to calculate the hash value as a function of contents of the object. Since you consider both the entry and num to determine the hashcode value of the object, these two objects has distinct hashcodes since they are different from entry. Thus they belong in two different hash buckets and never be recognized as the same.
However, if you set the entry as follows
k2.entry = "One";
then both k1 and k2 has the same hashcode value. Then they both belong to the same hash bucket and according to the equals method two objects are the same. Therefore, now the duplicates are ignored.
But, there's a problem lurking here. Ideally, equal objects must have equal hash codes. But, in your case it is not. Two objects are equal if they have the same num value, but they may produce different hashcode values. So the correct solution would be to just change your hashCode as follows.
#Override
public int hashCode() {
return Integer.hashCode(num);
}
Now, it should behave the way you would expect without the hack we added by setting k2.entry = "One";

hashCode and equals methods must be consistent.
From documentation
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.
In your case despite they are equal their hashCodes are different.
When adding elements HashSet check hashe of the added object and only if there is existing object in set with the same hash, HashSet checks if these objects with same hashes are equal.

Do java objects used as hash keys need to be 'fully' immutable?

If hashCode() calculation uses immutable fields and equals() uses all the fields would it be a problem when the class is used as a hash key? E.g.
import java.util.Objects;
public class Car {
protected final long vin;
protected String state;
protected String plateNumber;
public Car( long v, String s, String p ) {
vin = v; state = s; plateNumber = p;
}
public void move( String s, String p ) {
state = s; plateNumber = p;
}
public int hashCode() {
return (int)( vin % Integer.MAX_VALUE );
}
public boolean equals( Object other ) {
if (this == other) return true;
else if (!(other instanceof Car)) return false;
Car otherCar = (Car) other;
return vin == otherCar.vin
&& Objects.equals( state, otherCar.state )
&& Objects.equals( plateNumber, otherCar.plateNumber );
}
}
And move() is called on a car object after it is inserted into a hashset, possible via a reference kept elsewhere.
I am not after performance issues here. Only correctness.
I have read java hashCode contact, few answers on SO including this by venerable Jon Skeet and this from big blue. I feel that the last link gives the best explanation and imply that above code is correct.
Edit
Conclusion:
This class satisfy constraints placed on ‘equals()’ and ‘hashCode()’ in java. However it violates restrictions additional requirements placed on ‘equals()’ when used as keys in collections, hashed or not.
The additional requirement is that ‘equals()’ need to be consistent as long as the object is a key.
See the counter example by Louis Wasserman and the reference provided by Douglas below.
Few clarifications:
A) This class satisfy java object level constraints:
( carA == carB ) implies ( carA.hashCode() == carB.hashCode() )
( carA.hashCode() != carB.hashCode() ) implies ( carA != carB )
equals() need to be reflexive, symmetric, transitive.
hashCode() need to be consistent. i.e. Cannot change for an object during its lifetime.
equals() need to be consistent as long as neither object is modified.
Note that the reverse of ‘1.’ and ‘2.’ are not necessary. And the class above satisfies all the conditions.
Also java docs mention "equals() … implements the most discriminating possible equivalence relation on objects", but not sure if that is compulsory.
B) As for performance, the increment in collision avoidance probability decrease with each successive member variable we combine. Usually few well chosen member variables is sufficient.

It's correct if you never, ever call move after the Car is in the map. Otherwise it's wrong. Both hashCode and equals have to stay consistent after a key is in the map.

When considering only the hashCode and equals contracts, you are correct that this implementation satisfies their requirements. hashCode using a strict subset of the fields that equals uses is sufficient to guarantee that a.equals(b) implies a.hashCode() == b.hashCode() as required.
However, things change when you bring in Map. From the Map javadoc, "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."
After you call move on a Car that is a key in a Map, the behavior of that Map is now unspecified. In many cases it will in practice still work the way you want it to, but bizarre things could happen in ways that are hard to predict. While it would technically be valid for the Map to spontaneously empty itself or switch all lookups to use a random number generator, a more likely scenario might go like this:
Car car1 = ...
Car car2 = ... // a copy of car1
Map<Car, String> map1 = ...
map1.put(car1, "value");
assert map1.get(car2).equals("value"); // true
car1.move(...);
assert map1.get(car2).equals("value"); // NullPointerException on the equals call, car2 is no longer found
Notice that neither car2 nor the Map were changed themselves in any way, but the mapping of car2 changed (or rather, disappeared) anyway. This behavior is not officially specified, but I would guess most Map implementations do behave this way.

You may mutate your key candidates as much as you want, before or after (not during) they are used as keys.
In practice, it is very hard to enforce this rule. If you mutate objects you do not have a control if somebody uses them as keys or not.
Immutability for keys is just easier, removes source of subtle, hard-to-find bugs and just work better for key.
In your case I see no correctness issues. But why you ever bother not to include all fields in hashcode?

Short answer: it should be OK, but prepare for bizarre behavior.
Longer answer: when you change fields that participate in equals() on a key, the value keyed by that key will no longer be found.
Still longer answer: this looks as X/Y problem: you're asking about X, but you really need X to accomplish Y. Maybe you should ask about Y?
The car in your case is uniquely identified by vin. A car equals to itself. But, a car can be registered in different states. Maybe the answer is to have a Registration object (or a few of them) attached to the car? And then you can separate car.equals() from registration.equals().

Hash works by putting items into "buckets". Each bucket is calculated by the hashcode. After finding the bucket then the search continues comparing each item one by one using equals.
For example:
During insertion: an object whose id is 100 is placed in bucket 5 (the hashcode calculated 5).
During retrieval: you ask the hashmap to find the item 100. If the hash calculates 7 now then the algorithm will search for your object in bucket 7 but your object will never be found as it is dwelling in bucket 5.
In summary: the hash code and the actual key work together. The former is used to know in which bucket the item should be. The latter is used by the equals comparison seeking the actual item to return.

When your hashCode() implementation uses only limited number of fields (vs equals) you're reducing performance of almost any algorithm/data structure that uses hashing: HashMap, HashSet etc. You're increasing collision probability - it's the situation when two different objects (equals return false) have the same hash value.

The short answer is: No.
Long answer:
Fully immutability is not neccessary. BUT:
Equals must only depend on immutable values. Hashcode must depend on immutable values either a constant or a subset of the values used in equals or all values used in equals. Values that are not mentioned within equals mustn't be part of hashcode.
If you mutate values equals and hashcode rely on it is likely that you do not find your objects again in a hash based datastructure. Look at this:
public class Test {
private static class TestObject {
private String s;
public TestObject(String s) {
super();
this.s = s;
}
public void setS(String s) {
this.s = s;
}
#Override
public boolean equals(Object obj) {
boolean equals = false;
if (obj instanceof TestObject) {
TestObject that = (TestObject) obj;
equals = this.s.equals(that.s);
}
return equals;
}
#Override
public int hashCode() {
return this.s.hashCode();
}
}
public static void main(String[] args) {
TestObject a1 = new TestObject("A");
TestObject a2 = new TestObject("A");
System.out.println(a1.equals(a2)); // true
HashMap<TestObject, Object> hashSet = new HashMap<>(); // hash based datastructure
hashSet.put(a1, new Object());
System.out.println(hashSet.containsKey(a1)); // true
a1.setS("A*");
System.out.println(hashSet.containsKey(a1)); // false !!! // Object is not found as the hashcode used initially before mutation was used to determine the hash bucket
a2.setS("A*");
System.out.println(hashSet.containsKey(a2)); // false !!! Because a1 is in wrong hash bucket ...
System.out.println(a1.equals(a2)); // ... even if the objects are equals
}
}

default implementation of hashcode returns different values for objects constructed the same way

Here I am writing one sample code:
public class Test {
private int i;
private int j;
public Test() {
// TODO Auto-generated constructor stub
}
public Test(int i, int j)
{
this.i=i;
this.j=j;
}
}
now I am creating two objects as bellow:
Test t1= new Test(4,5);
Test t2 = new Test(4,5);
But when i am printing t1.hashcode() and t2.hashcode() they are giving different values.
But as per java's general contact they should return same value.
In fact, when i am doing same thing with String or Integer they are returning same hashcode(). Can anyone please explain why hashcode is different for t1 and t2 object?

But as per java's general contact they should return same value.
Java's equals-hashCode contract requires that if two objects are equal by Object.equals, they must have the same hashcode from Object.hashCode. But the default implementation of Object.equals is reference equality, and therefore two instances are the same if and only if they are the same instance.
Therefore, in particular, your two instances t1 and t2 are in fact not equal because you have not overridden Object.equals. They are not equal as references, and therefore not equal per Object.equals, and therefore it is acceptable for hashCode to possibly return different values. In fact, the contract explicitly says the following:
It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method on each of the two objects must produce distinct integer results.
Thus, we do not have a violation of the equals-hashCode contract here.
So, for your objects, if you want different instances to be equal per a logical definition of equality, you need to override Object.equals:
#Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
if (this == obj) {
return true;
}
if (!(obj instanceof Test)) {
return false;
}
Test other = (Test)obj;
return this.i == other.i && this.j == other.j;
}
And the equals-hashCode contract requires that you override Object.hashCode too or you'll run into some nasty bugs:
#Override
public int hashCode() {
int hash = 17;
hash = 31 * hash + this.i;
hash = 31 * hash + this.j;
return hash;
}
What does the contract say:
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.
Let's see if we have satisfied this requirement here. If x and y are instances of Test and satisfy x.equals(y) is true, we have that x.i == y.i and x.j == y.j. Then, clearly, if we invoke x.hashCode() and y.hashCode() we have the invariant that at each line of execution in Test.hashCode we will have hash holding the same value. Clearly this is true on the first line since hash will be 17 in both cases. It will hold on the second line since this.i will return the same value whether this == x or this == y because x.i equals y.i. Finally, on the penultimate line, we will still have hash being equal across both invocations because x.j equals y.j is true as well.
Note that there is one last piece of the contract that we haven't discussed yet. This is the requirement that hashCode return a consistent value during a single execution of a Java application:
Whenever it is invoked on the same object more than once during an execution of a Java application, the hashCode method must consistently return the same integer, provided no information used in equals comparisons on the object is modified.
The necessity of this is obvious. If you change the return value from hashCode during a single execution of the same application, you could lose your objects in hashtable-like data structures that use hashCode to keep track of objects. In particular, this is why mutating objects that are keys in hashtable-like data structures is pure evil; don't do it. I would go so far as to argue that they should be immutable objects.
In fact, when i am doing same thing with String or Integer they are returning same hashcode().
They've both overridden Object.equals and Object.hashCode.

You have not overridden the equals method in your class so the default one will be used that belongs to Object class.
Object class methods simply checks for the references whether they are referring to the same object or not.
Test t1 = new Test(4,5);
Test t2 = new Test(4,5);
are two different objects, if you don't override the equals method here, they will be equal if and only if you do
Test t2 = t1;
As you are creating two different objects here, hashcode which are NOT equal because they don't refer to the same object, hashcodes must be differnt
Remember
If two objects are equal, then their hashcode MUST be equal
But if hashcodes are equal, then its not necessary that objects should be equal

This is because of the default implementation of equals and hashCode in Java.
The JVM has no way of knowing how you decide that two objects are the same. What it does is use memory references. So, by default, the equals and hashCode methods compare memory references; i.e. two different objects are never .equals.
If you want to override this behaviour (and it is recommended you do so if you wish to use Collections for example) then all you need to do is implement your own equals and hashCode methods.

The problem is that t1 and t2 are not the same object they are different object. All objects created with new are different objects. And the default hashCode() implementation usually returns different hash codes for different objects. See Object.hashCode API.

Hashset allows duplicates?

This question surely isn't a new one, but I didn't find any helpful answer anywhere.
As you can see in the code below, the equals and hashcode methods are overriden, but it still allows duplicates. The Hashcode has been generated automatically by Netbeans.
#Override
public boolean equals(Object o)
{
TaskDetails other = (TaskDetails) o;
if ( (id_subtask == other.id_subtask)
&& ((date.compareTo(other.date)) == 0) )
{
System.err.println("Duplicate Entry"+id_subtask+" + "+other.id_subtask);
return true;
}
else
{
System.out.println("Good!" +id_subtask+" + "+other.id_subtask);
return false;
}
}
#Override
public int hashCode() {
int hash = 7;
hash = 71 * hash + this.id_subtask;
hash = 71 * hash + this.id_team_member;
hash = 71 * hash + Float.floatToIntBits(this.nb_hours);
hash = 71 * hash + (this.date != null ? this.date.hashCode() : 0);
hash = 71 * hash + (this.comment != null ? this.comment.hashCode() : 0);
hash = 71 * hash + (this.subtask_name != null ? this.subtask_name.hashCode() : 0);
System.out.println("Hash : "+hash + "Subtask : " + id_subtask);
return hash;
}
This the code used to add an entry into the hashset :
TaskDetails newTaskDetails = new TaskDetails
(
s.getId_subtask(),
mus.teamMember.getId_team_member(),
f,
mysqlFormat.format(caldate),
c.substring(0, Math.min(c.length(), 100)),
s.getName_subtask()
);
allTasks.add(newTaskDetails);
(allTasks being the Hashset)
This code is used in function A and B.
If only function A is executed, it works fine. If function B is executed after function A (so the code above is executed twice), then the hashset suddenly accepts duplicates, even though system.err is triggered saying there is a duplicate entry?
Is there a flaw in the code, or am I just missing something?
Thanks for the help!

you are using 2 fields to consider 2 objects to be "equal", but you are using more than 2 fields to construct the hashcode. your hashCode() method cannot be more specific than your equals() method. as a good rule of thumb, your hashCode() method should not use any fields that your equals() method does not use (it can use fewer however). to put it more technically, if 2 objects are "equal" they must have the same hashcode (the reverse is not required).

You are violating the consistency requirement between hashCode() and equals(). If two objects are equal according to equals(), they must also have the same hash. Because your equals only considers two fields, and hashCode considers more, this requirement is not met.

Your problem is that the implementation of hashCode() does not match equals(). Both methods must use the same attributes of your object.
It's likely in your implementation that the hashCode() is different even if equals() evaluates to true. In this case (different hashCodes) the objects are different for the HashMap.
Please correct your implementations to use the same attributes. Then the error should vanish.

From the javadoc of Object
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.
Your hashcodes are different for two objects that are equal according to the equals(Object) method, so the other code HashSet is going to make the wrong assumptions, and return the wrong results.
Some code is written in a manner that they depend on other objects honoring "contracts". Your class doesn't honor the Object contract, so nothing in collections can be assumed to work, as collections requires that the Object contracts not be broken.

This is a duplicate question, see my previous answer.
The behaviour where a java.util.HashSet allows duplicates is caused when the hash code of the objects in the java.util.HashSet can change.
This typically happens when an object's hash code is constructed from mutable fields.

In Java, why must equals() and hashCode() be consistent?

If I override either method on a class, it must make sure that if A.equals(B) == true then A.hashCode() == B.hashCode must also be true.
Can someone show me a simple example where if this is violated, it'll cause a problem? I think it has something to do with if you use that class as the type of keys to Hashmap?

Sure:
public class Test {
private final int m, n;
public Test(int m, int n) {
this.m = m;
this.n = n;
}
public int hashCode() { return n * m; }
public boolean equals(Object ob) {
if (ob.getClass() != Test.class) return false;
Test other = (Test)ob;
return m == other.m;
}
}
with:
Set<Test> set = new HashSet<Test>();
set.put(new Test(3,4));
boolean b = set.contains(new Test(3, 10)); // false
Technically that should be true because m == 3 in both cases.
In general a HashMap works like this: it has a variable number of what are commonly called "buckets". The number of buckets can change over time (as entries are added and removed) but it is always a power of 2.
Let's say a given HashMap has 16 buckets. When you call put() to add an entry, the hashCode() of the key is calculated and then a mask is taken depending on the size of the buckets. If you (bitwise) AND the hashCode() with 15 (0x0F) you will get the last 4 bits, equaling a number between 0 and 15 inclusive:
int factor = 4;
int buckets = 1 << (factor-1) - 1; // 16
int mask = buckets - 1; // 15
int code = key.hashCode();
int dest = code & mask; // a number from 0 to 15 inclusive
Now if there is already an entry in that bucket you have what's called a collision. There are multiple ways of dealing with this but the one used by HashMap (and is probably the most common overall) is bucketing. All the entries with the same masked hashCode are put in a list of some kind.
So to find if a given key is in the map already:
Calculate the masked hash code;
Find the appropriate bucket;
If it's empty, key not found;
If is isn't empty, loop through all entries in the bucket checking equals().
Looking through a bucket is a linear (O(n)) operation but it's on a small subset. The hashcode bucket determination is essentially constant (O(1)). If buckets are sufficiently small then access to a HashMap is usually described as "near O(1)".
You can make a couple of observations about this.
Firstly, if you have a bunch of objects that all return 42 as their hash code a HashMap will still work but it will operate as an expensive list. Access will be O(n) (as everything will be in the same bucket regardless of the number of buckets). I've actually been asked this in an interview.
Secondly, returning to your original point, if two objects are equal (meaning a.equals(b) == b.equals(a) == true) but have different hash codes then the HashMap will go looking in (probably) the wrong bucket resulting in unpredictable and undefined behaviour.

This is discussed in the Item 8: Always override hashCode when you override equals of Joshua Bloch's Effective Java:
A common source of bugs is the failure to override the hashCode method. You must
override hashCode in every class that overrides equals. Failure to do so will
result in a violation of the general contract for Object.hashCode, which will pre-
vent your class from functioning properly in conjunction with all hash-based collec-
tions, including HashMap, HashSet, and Hashtable.
Here is the contract, copied from the
java.lang.Object specification:
Whenever it is invoked on the same object more than once during an execution of an application, the hashCode method must consistently return the same integer, provided no information used in equals comparisons on the object is modified. This integer need not remain consistent from one execution of an application to another execution of the same application.
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.
It is not required that if two objects are unequal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce distinct integer results. However, the programmer should be aware that producing distinct integer results for unequal objects may improve the performance of hash tables.
The key provision that is violated when you fail to override hashCode is
the second one: Equal objects must have equal hash codes. Two distinct
instances may be logically equal according to the class’s equals method, but to
the Object class’s hashCode method, they’re just two objects with nothing much
in common. Therefore object’s hashCode method returns two seemingly random
numbers instead of two equal numbers as required by the contract.
For example, consider the following simplistic PhoneNumber class, whose
equals method is constructed according to the recipe in Item 7:
public final class PhoneNumber {
private final short areaCode;
private final short exchange;
private final short extension;
public PhoneNumber(int areaCode, int exchange,
int extension) {
rangeCheck(areaCode, 999, "area code");
rangeCheck(exchange, 999, "exchange");
rangeCheck(extension, 9999, "extension");
this.areaCode = (short) areaCode;
this.exchange = (short) exchange;
this.extension = (short) extension;
}
private static void rangeCheck(int arg, int max,
String name) {
if (arg < 0 || arg > max)
throw new IllegalArgumentException(name +": " + arg);
}
public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof PhoneNumber))
return false;
PhoneNumber pn = (PhoneNumber)o;
return pn.extension == extension &&
pn.exchange == exchange &&
pn.areaCode == areaCode;
}
// No hashCode method!
... // Remainder omitted
}
Suppose you attempt to use this class
with a HashMap:
Map m = new HashMap();
m.put(new PhoneNumber(408, 867, 5309), "Jenny");
At this point, you might expect
m.get(new PhoneNumber(408 , 867,
5309)) to return "Jenny", but it
returns null. Notice that two PhoneNumber instances are
involved: One is used for insertion
into the HashMap, and a second, equal,
instance is used for (attempted)
retrieval. The PhoneNumber class’s
failure to override hashCode causes
the two equal instances to have
unequal hash codes, in violation of
the hashCode contract. Therefore the
get method looks for the phone number
in a different hash bucket from the
one in which it was stored by the put
method. Fixing this problem is as
simple as providing a proper hashCode
method for the PhoneNumber class.
[...]
See the Chapter 3 for the full content.

Containers like HashSet rely on the hash function to determine where to put it, and where to get it from when asked for it. If A.equals(B), then a HashSet is expecting A to be in the same place as B. If you put A in with value V, and look up B, you should expect to get V back (since you've said A.equals(B)). But if A.hashcode() != B.hashcode(), then the hashset may not find where you put it.

Here's a little example:
Set<Foo> myFoos = new HashSet<Foo>();
Foo firstFoo = new Foo(123,"Alpha");
myFoos.add(firstFoo);
// later in the processing you get another Foo from somewhere
Foo someFoo = //use imagination here...;
// maybe you get it from a database... and it's equal to Foo(123,"Alpha)
if (myFoos.contains(someFoo)) {
// maybe you win a million bucks.
}
So, imagine that the hashCode that gets created for firstFoo is 99999 and it winds up at a specific spot in the myFoos HashSet. Later when you get the someFoo and you look for it in the myFoos HashSet, it needs to generate the same hashCode so you can find it.

It's exactly because of hash tables.
Because of the possibility of hash code collisions, hash tables need to check identity as well, otherwise the table can't determine if it found the object it was looking for, or one with the same hash code. So every get() in a hash table calls key.equals(potentialMatch) before returning a value.
If equals() and hashCode() are inconsistent you can get very inconsistent behavior. Say for two objects, a and b, a.equals(b) returns true, but a.hashCode() != b.hashCode(). Insert a and a HashSet will return false for .contains(b), but a List created from that set will return true (because the list doesn't use hash codes).
HashSet set = new HashSet();
set.add(a);
set.contains(b); // false
new ArrayList(set).contains(b); // true
Obviously, that could be bad.

The idea behind this is that two objects are "equal" if all of their fields have equal values. If all of fields have equal values, the two objects should have the same hash value.