This question already has answers here:
Java HashSet contains duplicates if contained element is modified
(7 answers)
Mutable objects and hashCode
(6 answers)
What happens to the lookup in a Hashmap or Hashset when the objects Hashcode changes
(4 answers)
Java: Modify id that changes hashcode
(1 answer)
mutable fields for objects in a Java Set
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Closed 5 years ago.
Consider the employee class -
public class Employer implements Serializable{
private Long id;
private String name;
#Override
public boolean equals(Object obj) {
if (obj == null)
return false;
if (obj instanceof Employer) {
Employer employer = (Employer) obj;
if (this.id == employer.id) {
return true;
}
}
return false;
}
//Idea from effective Java : Item 9
#Override
public int hashCode() {
int result = 17;
result = 31 * result + id.hashCode();
//result = 31 * result + name.hashCode();
return result;
}
}
With 2 employee objects created -
Employer employer1 = new Employer();
employer1.setId(10L);
Employer employer2 = new Employer();
employer2.setId(11L);
After adding them to the hashset, the size will be 2.
HashSet internally uses a hashmap to maintain the uniqueness-
private transient HashMap<E,Object> map;
public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
Now, if I set the id for the second employee to be same as that of the first, i.e-
employer2.setId(10L);
the size still remains 2.
Why is it not 1? Does the in-variants get destroyed?
All hash-based containers, including HashSet<T>, make a very important assumption about hash code of their keys: they assume that hash code never changes while the object is inside the container.
Your code violates this assumption by modifying the instance while it is still in the hash set. There is no practical way for HashSet<T> to react to this change, so you must pick one of two ways to deal with this issue:
Never modify keys of hash-based containers - This is by far the most common approach, often achieved by making hash keys immutable.
Keep track of modifications, and re-hash objects manually - essentially, your code makes sure that all modifications to hash keys happen while they are outside containers: you remove the object from the container, make modifications, and then put it back.
The second approach often becomes a source of maintenance headaches. When you need to keep mutable data in a hash-based container, a good approach is to use only final fields in the computation of your hash code and equality checks. In your example this would mean making id field final, and removing setId method from the class.
the size still remains 2. Why is it not 1? Does the in-variants get destroyed?
If you modify any of the properties used to compute hashCode and equals for an instance already in the HashSet, the HashSet implementation is not aware of that change.
Therefore it will keep the two instances, even though they are now equal to each other.
You shouldn't make such updates for instances that are members or HashSets (or keys in HashMaps). If you must make such changes, remove the instance from the Set before mutating it and re-add it later.
Related
In Java, obj.hashCode() returns some value. What is the use of this hash code in programming?
hashCode() is used for bucketing in Hash implementations like HashMap, HashTable, HashSet, etc.
The value received from hashCode() is used as the bucket number for storing elements of the set/map. This bucket number is the address of the element inside the set/map.
When you do contains() it will take the hash code of the element, then look for the bucket where hash code points to. If more than 1 element is found in the same bucket (multiple objects can have the same hash code), then it uses the equals() method to evaluate if the objects are equal, and then decide if contains() is true or false, or decide if element could be added in the set or not.
From the Javadoc:
Returns a hash code value for the object. This method is supported for the benefit of hashtables such as those provided by java.util.Hashtable.
The general contract of hashCode is:
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. 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(java.lang.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 hashtables.
As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. (This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the Java programming language.)
hashCode() is a function that takes an object and outputs a numeric value. The hashcode for an object is always the same if the object doesn't change.
Functions like HashMap, HashTable, HashSet, etc. that need to store objects will use a hashCode modulo the size of their internal array to choose in what "memory position" (i.e. array position) to store the object.
There are some cases where collisions may occur (two objects end up with the same hashcode), and that, of course, needs to be solved carefully.
The value returned by hashCode() is the object's hash code, which is the object's memory address in hexadecimal.
By definition, if two objects are equal, their hash code must also be equal. If you override the equals() method, you change the way two objects are equated and Object's implementation of hashCode() is no longer valid. Therefore, if you override the equals() method, you must also override the hashCode() method as well.
This answer is from the java SE 8 official tutorial documentation
A hashcode is a number generated from any object.
This is what allows objects to be stored/retrieved quickly in a Hashtable.
Imagine the following simple example:
On the table in front of you. you have nine boxes, each marked with a number 1 to 9. You also have a pile of wildly different objects to store in these boxes, but once they are in there you need to be able to find them as quickly as possible.
What you need is a way of instantly deciding which box you have put each object in. It works like an index. you decide to find the cabbage so you look up which box the cabbage is in, then go straight to that box to get it.
Now imagine that you don't want to bother with the index, you want to be able to find out immediately from the object which box it lives in.
In the example, let's use a really simple way of doing this - the number of letters in the name of the object. So the cabbage goes in box 7, the pea goes in box 3, the rocket in box 6, the banjo in box 5 and so on.
What about the rhinoceros, though? It has 10 characters, so we'll change our algorithm a little and "wrap around" so that 10-letter objects go in box 1, 11 letters in box 2 and so on. That should cover any object.
Sometimes a box will have more than one object in it, but if you are looking for a rocket, it's still much quicker to compare a peanut and a rocket, than to check a whole pile of cabbages, peas, banjos, and rhinoceroses.
That's a hash code. A way of getting a number from an object so it can be stored in a Hashtable. In Java, a hash code can be any integer, and each object type is responsible for generating its own. Lookup the "hashCode" method of Object.
Source - here
Although hashcode does nothing with your business logic, we have to take care of it in most cases. Because when your object is put into a hash based container(HashSet, HashMap...), the container puts/gets the element's hashcode.
hashCode() is a unique code which is generated by the JVM for every object creation.
We use hashCode() to perform some operation on hashing related algorithm like Hashtable, Hashmap etc..
The advantages of hashCode() make searching operation easy because when we search for an object that has unique code, it helps to find out that object.
But we can't say hashCode() is the address of an object. It is a unique code generated by JVM for every object.
That is why nowadays hashing algorithm is the most popular search algorithm.
One of the uses of hashCode() is building a Catching mechanism.
Look at this example:
class Point
{
public int x, y;
public Point(int x, int y)
{
this.x = x;
this.y = y;
}
#Override
public boolean equals(Object o)
{
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Point point = (Point) o;
if (x != point.x) return false;
return y == point.y;
}
#Override
public int hashCode()
{
int result = x;
result = 31 * result + y;
return result;
}
class Line
{
public Point start, end;
public Line(Point start, Point end)
{
this.start = start;
this.end = end;
}
#Override
public boolean equals(Object o)
{
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Line line = (Line) o;
if (!start.equals(line.start)) return false;
return end.equals(line.end);
}
#Override
public int hashCode()
{
int result = start.hashCode();
result = 31 * result + end.hashCode();
return result;
}
}
class LineToPointAdapter implements Iterable<Point>
{
private static int count = 0;
private static Map<Integer, List<Point>> cache = new HashMap<>();
private int hash;
public LineToPointAdapter(Line line)
{
hash = line.hashCode();
if (cache.get(hash) != null) return; // we already have it
System.out.println(
String.format("%d: Generating points for line [%d,%d]-[%d,%d] (no caching)",
++count, line.start.x, line.start.y, line.end.x, line.end.y));
}
When I edit an object, which is contained within a HashSet, the hash of the object changes, but the HashSet is not updated internally. Therefor, I practically can add the same object twice:
TestObject testObject = new TestObject(1, "hello");
Set<TestObject> set = new HashSet<>();
set.add(testObject);
testObject.number = 2;
set.add(testObject);
set.forEach(System.out::println);
//will print
//{number:2, string:hello}
//{number:2, string:hello}
Full working code example:
import java.util.*;
public class Main {
public static void main(String[] args) {
TestObject testObject = new TestObject(1, "hello");
Set<TestObject> set = new HashSet<>();
// add initial object
set.add(testObject);
// modify object
testObject.number = 2;
testObject.string = "Bye";
// re-add same object
set.add(testObject);
set.forEach(System.out::println);
}
}
class TestObject {
public int number;
public String string;
public TestObject(int number, String string) {
this.number = number;
this.string = string;
}
#Override
public int hashCode() {
return Objects.hash(number, string);
}
#Override
public boolean equals(Object obj) {
if (!(obj instanceof TestObject)) {
return false;
}
TestObject o = (TestObject) obj;
return number == o.number && string.equals(o.string);
}
#Override
public String toString() {
return "{number:" + number + ", string:" + string + "}";
}
}
This means, after modifying an object which already is contained in a HashSet, theHashSet` turns unreliable or invalid.
Modifying an object that is somewhere contained in a Set (probably even without knowing) seems a regular use case to me . And something which I probably already have done a lot.
This throws me back and brings one basic question to me: When or why should I use a HashSet if it has such a behaviour?
Well, if you have a look at the HashSet source you'll see that it's basically a HashMap<E, Object> with the elements being the keys - and modifying keys of a hashmap is never a good idea. The map/set will not be updated if the hash would change, in fact the map/set wouldn't even know about that change.
In general keys of a HashMap or elements in a HashSet should be immutable in that their hash and equality doesn't change. In most cases the hash and equality are based on those object's (business) identity, so if number and string are both part of that object's identity then you shouldn't be able to change those.
Modifying an object that is somewhere contained in a Set (probably even without knowing) seems a regular use case to me . And something which I probably already have done a lot.
It's probably true that objects contained in sets are modified quite often but that normally would mean that data that's not used to generate the hashcode or to check equality are modified. As an example let's say a person's hashcode is based on their ID number. That would mean that hashCode() and equals() should only be based on that number and that everything else could be modified safely.
So you could modify elements in a HashSet as long as you're not modifying their "id".
When or why should I use a HashSet if it has such a behaviour?
If you need to store mutable objects in a HashSet you have a few options which basically revolve around using only the immutable parts for hashCode() and equals(). For sets that could be done by using a wrapper object that provides a customized implementation for those methods. Alternatively you could extract one or more immutable properties and use those as the key into a map (in case of multiple properties you'd need to build some sort of key object out of those)
You’re never supposed to compare strings with == use .equals instead
Adding an element that is already present, as you said, won't override the element that is already in the HashSet. Use a remove(), before calling the add(), to insure the new value to be inserted effectively.
Side note: as some users have noted, pay attention to the Strings' comparisons in your test.
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
}
}
I have a class that contains a collection.
The two instances of the class are equal if the contents of the collection are equal.
While I am building my data structure I store the class in a HashSet, and the contents of the collection change. The changes cause a change in the hash code value. This seems to cause side effects where my data is lost in the Set.
Removing the collection from the hashcode calculation fixes the problem, but violates rule where all fields in equals should be used in the hashcode.
How would you implement the hashcode in this situation?
public class LeveZeroHolder
{
private final Set<LevelOneHolder> orgGroups = new HashSet<LevelOneHolder>();
private final String name;
public LeveZeroHolder(String name, LevelOneHolder og)
{
this.name = name;
orgGroups.add(og);
og.setFA(this);
}
#Override
public boolean equals(Object obj)
{
if (this == obj)
return true;
if (obj == null || obj.getClass () != getClass ())
return false;
LeveZeroHolder hobj = (LeveZeroHolder)obj;
return getOrgGroups().equals(hobj.getOrgGroups()) && getName().equals(hobj.getName());
}
#Override
public int hashCode()
{
int rs = 17;
rs = rs * 37 + ((getName() == null) ? 0 : getName().hashCode ());
rs = rs * 37 + ((getOrgGroups() == null) ? 0 : getOrgGroups().hashCode());
return rs;
}
public String getName()
{
return name;
}
public Set<LevelOneHolder> getOrgGroups()
{
return orgGroups;
}
public void addOrgGroup(LevelOneHolder o)
{
o.setFA(this);
orgGroups.add(o);
}
}
If you mean that when you store such objects as a key in a Map or in a Set they get lost, you might want to have a look at this thread which explains why storing mutable objects in a set in not a good idea, especially if they change while being held by the set.
Extract from the Set javadoc:
Note: Great care must be exercised if mutable objects are used as set elements. The behavior of a set is not specified if the value of an object is changed in a manner that affects equals comparisons while the object is an element in the set. A special case of this prohibition is that it is not permissible for a set to contain itself as an element.
If your collection of LevelOneHolder really does affect the uniqueness of your object then you should make LevelZeroHolder immutable. If you add a LevelOneHolder to your LevelZeroHolder then you should, instead of updating the collection, return a completely new LevelZeroHolder with the collection copied from the pre-existing one and joined with the new one you want to add.
In this way the hashCode never changes but you end up with different LevelZeroHolders with different hashCodes. This is probably correct though considering that you're saying that the nested collection within LevelZeroHolder contributes to the uniqueness of that object.
If you put your object in a data structure where you depend on the hashCode or the equals() to find it, AND if you change some property of the object (like its name, or something in the list in it) when it is already in that data structure, then you wont find it. Is that your case?
The reason is that when you put the object in such a data structure, it is inserted according to its hashCode, at the time of insertion. If when its inside, you change something that changes the hashCode, then your object is still stored according to the old hashCode but when you try to get it, and so you compute its hashCode to see if thats your object, then you get the new hashCode value and so you dont see that its the one you want.
The rule is that for an object in a data structure that uses the object's hash code or equals NEVER change the object in a way that affects the hashCode or equals.
Try using immutable objects instead.
Have the following class:
public class Member {
private int x;
private long y;
private double d;
public Member(int x, long y, double d) {
this.x = x;
this.y = y;
this.d = d;
}
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + x;
result = (int) (prime * result + y);
result = (int) (prime * result + Double.doubleToLongBits(d));
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (obj instanceof Member) {
Member other = (Member) obj;
return other.x == x && other.y == y
&& Double.compare(d, other.d) == 0;
}
return false;
}
public static void main(String[] args) {
Set<Member> test = new HashSet<Member>();
Member b = new Member(1, 2, 3);
test.add(b);
System.out.println(b.hashCode());
b.x = 0;
System.out.println(b.hashCode());
Member first = test.iterator().next();
System.out.println(test.contains(first));
System.out.println(b.equals(first));
System.out.println(test.add(first));
}
}
It produces the following results:
30814
29853
false
true
true
Because the hashCode depends of the state of the object it can no longer by retrieved properly, so the check for containment fails. The HashSet in no longer working properly. A solution would be to make Member immutable, but is that the only solution? Should all classes added to HashSets be immutable? Is there any other way to handle the situation?
Regards.
Objects in hashsets should either be immutable, or you need to exercise discipline in not changing them after they've been used in a hashset (or hashmap).
In practice I've rarely found this to be a problem - I rarely find myself needing to use complex objects as keys or set elements, and when I do it's usually not a problem just not to mutate them. Of course if you've exposed the references to other code by this time, it can become harder.
Yes. While maintaining your class mutable, you can compute the hashCode and the equals methods based on immutable values of the class ( perhaps a generated id ) to adhere to the hashCode contract defined in Object class:
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. 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(java.lang.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 hashtables.
Depending on your situation this may be easier or not.
class Member {
private static long id = 0;
private long id = Member.id++;
// other members here...
public int hashCode() { return this.id; }
public boolean equals( Object o ) {
if( this == o ) { return true; }
if( o instanceOf Member ) { return this.id == ((Member)o).id; }
return false;
}
...
}
If you need a thread safe attribute, you may consider use: AtomicLong instead, but again, it depends on how are you going to use your object.
As already mentioned, one can accept the following three solutions:
Use immutable objects; even when your class is mutable, you may use immutable identities on your hashcode implementation and equals checking, eg an ID-like value.
Similarly to the above, implement add/remove to get a clone of the inserted object, not the actual reference. HashSet does not offer a get function (eg to allow you alter the object later on); thus, you are safe there won't exist duplicates.
Exercise discipline in not changing them after they've been used, as #Jon Skeet suggests
But, if for some reason you really need to modify objects after being inserted to a HashSet, you need to find a way of "informing" your Collection with the new changes. To achieve this functionality:
You can use the Observer design pattern, and extend HashSet to implement the Observer interface. Your Member objects must be Observable and update the HashSet on any setter or other method that affects hashcode and/or equals.
Note 1: Extending 3, using 4: we may accept alterations, but those that do not create an already existing object (eg I updated a user's ID, by assigning a new ID, not setting it to an existing one). Otherwise, you have to consider the scenario where an object is transformed in such a way that is now equal to another object already existing in the Set. If you accept this limitation, 4th suggestion will work fine, else you must be proactive and define a policy for such cases.
Note 2: You have to provide both previous and current states of the altered object on your update implementation, because you have to initially remove the older element (eg use getClone() before setting new values), then add the object with the new state. The following snippet is just an example implementation, it needs changes based on your policy of adding a duplicate.
#Override
public void update(Observable newItem, Object oldItem) {
remove(oldItem);
if (add(newItem))
newItem.addObserver(this);
}
I've used similar techniques on projects, where I require multiple indices on a class, so I can look up with O(1) for Sets of objects that share a common identity; imagine it as a MultiKeymap of HashSets (this is really useful, as you can then intersect/union indices and work similarly to SQL-like searching). In such cases I annotate methods (usually setters) that must fireChange-update each of the indices when a significant change occurs, so indices are always updated with the latest states.
Jon Skeet has listed all alternatives. As for why the keys in a Map or Set must not change:
The contract of a Set implies that at any time, there are no two objects o1 and o2 such that
o1 != o2 && set.contains(o1) && set.contains(o2) && o1.equals(o2)
Why that is required is especially clear for a Map. From the contract of Map.get():
More formally, if this map contains a mapping from a key
k to a value v such that (key==null ? k==null : key.equals(k)), then this method returns v, otherwise it returns null. (There can be at most one such mapping.)
Now, if you modify a key inserted into a map, you might make it equal to some other key already inserted. Moreover, the map can not know that you have done so. So what should the map do if you then do map.get(key), where key is equal to several keys in the map? There is no intuitive way to define what that would mean - chiefly because our intuition for these datatypes is the mathematical ideal of sets and mappings, which don't have to deal with changing keys, since their keys are mathematical objects and hence immutable.
Theoretically (and more often than not practically too) your class either:
has a natural immutable identity that can be inferred from a subset of its fields, in which case you can use those fields to generate the hashCode from.
has no natural identity, in which case using a Set to store them is unnecessary, you could just as well use a List.
Never change 'hashable field" after putting in hash based container.
As if you (Member) registered your phone number (Member.x) in yellow page(hash based container), but you changed your number, then no one can find you in the yellow page any more.