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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));
}
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));
}
The HashSet class has an add(Object o) method, which is not inherited from another class. The Javadoc for that method says the following:
Adds the specified element to this set if it is not already present. More formally, adds the specified element e to this set if this set contains no element e2 such that (e==null ? e2==null : e.equals(e2)). If this set already contains the element, the call leaves the set unchanged and returns false.
In other words, if two objects are equal, then the second object will not be added and the HashSet will remain the same. However, I've discovered that this is not true if objects e and e2 have different hashcodes, despite the fact that e.equals(e2). Here is a simple example:
import java.util.HashSet;
import java.util.Iterator;
import java.util.Random;
public class BadHashCodeClass {
/**
* A hashcode that will randomly return an integer, so it is unlikely to be the same
*/
#Override
public int hashCode(){
return new Random().nextInt();
}
/**
* An equal method that will always return true
*/
#Override
public boolean equals(Object o){
return true;
}
public static void main(String... args){
HashSet<BadHashCodeClass> hashSet = new HashSet<>();
BadHashCodeClass instance = new BadHashCodeClass();
System.out.println("Instance was added: " + hashSet.add(instance));
System.out.println("Instance was added: " + hashSet.add(instance));
System.out.println("Elements in hashSet: " + hashSet.size());
Iterator<BadHashCodeClass> iterator = hashSet.iterator();
BadHashCodeClass e = iterator.next();
BadHashCodeClass e2 = iterator.next();
System.out.println("Element contains e and e2 such that (e==null ? e2==null : e.equals(e2)): " + (e==null ? e2==null : e.equals(e2)));
}
The results from the main method are:
Instance was added: true
Instance was added: true
Elements in hashSet: 2
Element contains e and e2 such that (e==null ? e2==null : e.equals(e2)): true
As the example above clearly shows, HashSet was able to add two elements where e.equals(e2).
I'm going to assume that this is not a bug in Java and that there is in fact some perfectly rational explanation for why this is. But I can't figure out what exactly. What am I missing?
I think what you're really trying to ask is:
"Why does a HashSet add objects with inequal hash codes even if they claim to be equal?"
The distinction between my question and the question you posted is that you're assuming this behavior is a bug, and therefore you're getting grief for coming at it from that perspective. I think the other posters have done a thoroughly sufficient job of explaining why this is not a bug, however they have not addressed the underlying question.
I will try to do so here; I would suggest rephrasing your question to remove the accusations of poor documentation / bugs in Java so you can more directly explore why you're running into the behavior you're seeing.
The equals() documentations states (emphasis added):
Note that it is generally necessary to override the hashCode method whenever this method is overridden, so as to maintain the general contract for the hashCode method, which states that equal objects must have equal hash codes.
The contract between equals() and hashCode() isn't just an annoying quirk in the Java specification. It provides some very valuable benefits in terms of algorithm optimization. By being able to assume that a.equals(b) implies a.hashCode() == b.hashCode() we can do some basic equivalence tests without needing to call equals() directly. In particular, the invariant above can be turned around - a.hashCode() != b.hashCode() implies a.equals(b) will be false.
If you look at the code for HashMap (which HashSet uses internally), you'll notice an inner static class Entry, defined like so:
static class Entry<K,V> implements Map.Entry<K,V> {
final K key;
V value;
Entry<K,V> next;
int hash;
...
}
HashMap stores the key's hash code along with the key and value. Because a hash code is expected to not change over the time a key is stored in the map (see Map's documentation, "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.") it is safe for HashMap to cache this value. By doing so, it only needs to call hashCode() once for each key in the map, as opposed to every time the key is inspected.
Now lets look at the implementation of put(), where we see these cached hashes being taken advantage of, along with the invariant above:
public V put(K key, V value) {
...
int hash = hash(key);
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))) {
// Replace existing element and return
}
}
// Insert new element
}
In particular, notice that the conditional only ever calls key.equals(k) if the hash codes are equal and the key isn't the exact same object, due to short-circuit evaluation. By the contract of these methods, it should be safe for HashMap to skip this call. If your objects are incorrectly implemented, these assumptions being made by HashMap are no longer true, and you will get back unusable results, including "duplicates" in your set.
Note that your claim "HashSet ... has an add(Object o) method, which is not inherited from another class" is not quite correct. While its parent class, AbstractSet, does not implement this method, the parent interface, Set, does specify the method's contract. The Set interface is not concerned with hashes, only equality, therefore it specifies the behavior of this method in terms of equality with (e==null ? e2==null : e.equals(e2)). As long as you follow the contracts, HashSet works as documented, but avoids actually doing wasteful work whenever possible. As soon as you break the rules however, HashSet cannot be expected to behave in any useful way.
Consider also that if you attempted to store objects in a TreeSet with an incorrectly implemented Comparator, you would similarly see nonsensical results. I documented some examples of how a TreeSet behaves when using an untrustworthy Comparator in another question: how to implement a comparator for StringBuffer class in Java for use in TreeSet?
You've violated the contract of equals/hashCode basically:
From the hashCode() docs:
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.
and from equals:
Note that it is generally necessary to override the hashCode method whenever this method is overridden, so as to maintain the general contract for the hashCode method, which states that equal objects must have equal hash codes.
HashSet relies on equals and hashCode being implemented consistently - the Hash part of the name HashSet basically implies "This class uses hashCode for efficiency purposes." If the two methods are not implemented consistently, all bets are off.
This shouldn't happen in real code, because you shouldn't be violating the contract in real code...
#Override
public int hashCode(){
return new Random().nextInt();
}
You are returning different has codes for same object every time it is evaluated. Obviously you will get wrong results.
add() function is as follows
public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
and put() is
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 you notice first has is calculated which is different in your case which is why object is added. equals() comes into picture only if hash are same for objects i.e collision has occured. Since in case hash are different equals() is never executed
if (e.hash == hash && ((k = e.key) == key || key.equals(k)))
Read more on what short circuiting is. since e.hash == hash is false nothing else is evaluated.
I hope this helps.
because hashcode() is really implemented very badly,
it will try to equate in each random bucket on each add(), if you return constant value from hashcode() it wouldn't let you enter any
It is not required that hash codes be different for all elements! It is only required that two elements are not equal.
HashCode is used first to find the hash bucket the object should occupy. If hadhcodes are different, objects are assumed to be not equal. If hashcodes are equal, then the equals() method is used to determine equality. The use of hashCode is an efficiency mechanism.
And...
Your hash code implementation violates the contract that it should not change unless the objects identifying fields change.
In the following code, the output shows CONTAINS for each object, whereas commenting out the anonymous object's equals() method results in MISSING, which leads me to believe the second equality pass (hashCode() -> equals()) actually calls the equality method of the supplied object instead of the object within the collection being tested.
List<String> strings = Arrays.asList("Hello", "there", "Qix");
HashSet<String> set = new HashSet<>(strings);
for(final String s : strings)
{
boolean contains = set.contains(new Object(){
#Override
public int hashCode() {
return s.hashCode();
}
#Override
public boolean equals(Object obj) {
return true;
}
});
System.out.format("%s: %s\n",
s,
contains ? "CONTAINS" : "MISSING");
}
Why is this? Is it because the equals() method, by principle, should be symmetric between the two objects?
The HashSet either has to do a.equals(b) or b.equals(a). And as they should be written to be symmetric*, it shouldn't matter which it chooses.
But for reference, the documentation states:
returns true if and only if this set contains an element e such that (o==null ? e==null : o.equals(e))
* See http://docs.oracle.com/javase/6/docs/api/java/lang/Object.html#equals(java.lang.Object).
It's an implementation detail which we shouldnt worry about since it's never said in public API how it uses equals. Like you said it's supposed to be symmetric anyway. If we go into src we'll see that it is really passedObject.equals(storedObject)
Becaue AFAIK the default Object.equals implementation uses reference comparison so it checks if the two references refer to the same object.
Here you are comparing instances of Strings with another custom classe, they can't be equal, whatever the way this is done.
Given two Lists, each list holding the same object type, I would like to find objects between the two lists that match, based on some property values.
e.g. an object from List1, L1Obj, matches an object from List2, L2Obj, if L1Obj.a == L2Obj.a AND L1Obj.b == L2Obj.c AND L1Obj.c == L2Obj.c
These properties are not the only properties of the of the class, but are all that is needed to uniquely identify an object within a list.
My question is - what is the best way to achieve this?
One way would be to construct to HashMaps based on the lists, with the concataned String value of a+b+c used as the key to index an object. That way I could iterate through the first list, and attempt to lookup an object in the second list with the same key.
How does this sound? Is there a better way of achieving this??
All help is much appreciated!
UPDATE:
Okay, so actually I need a bit more. Upon finding a match, I want to overwrite properties L1Obj.x, L1Obj.y, L1Obj.z with those of L2Obj. HashSet sounds great for finding that matches, but if I'm right it doesn't actually allow me to access these matches.
What can I do about this?
Do the objects you want to look at implement equals(Object) and hashCode() that only take into account the fields you care about? If so, you can create a new HashSet from the first list, and then call retainAll() passing in the second list.
If they don't implement equals(Object) and hashCode() with respect to the properties you care about, you can create a TreeSet and pass in a Comparator that looks at the properties you care about.
Rather than use the String repesntation, use the equals() method a HashSet as so:
class MyObj {
Property a;
Property b;
Property c;
public boolean equals(Object o) {
// use == if Property is primitive, like int or something
return o instanceof MyObj && a.equals(o.a) && b.equals(o.b) && c.equals(o.c);
}
// edit - when you override equals, also override hashcode
public int hashCode() {
return a.hashCode() ^ b.hashCode() ^ c.hashCode();
}
public String toString() {
return a.toString() + " " + b.toString() + " " + c.toString();
}
}
// later in your main method
Set<MyObj> objSet = new HashSet<MyObj>();
for(MyObj o : list1) objSet.add(o);
for(MyObj o : list2) if(objSet.contains(o)) System.out.println(o + " is a match!");
You can do one thing. Have two lists with these objects and override the equals method of the class to which these objects belong.
Your equals method should look like
#Override
public boolean equals(Object obj)
{
return (this.a == obj.a && this.b == obj.b && this.c == obj.c)
}
Also remember, once you override equals method, you need to override int hashCode() method as well.
One thing to note is while implementing hashCode() is that 2 equal objects will have same hashCode, while the converse is not true.
I don't know if I thinking to easy but I would try it like that:
Override the equals method of the object to implement your comparison to check if it is the same object
Then I would iterate over the first list and check with the contains method if the object is also contained in the second list.
Then I would iterate through the second list and check if the object is also in the first list and not already in the result list.
The object in question should implement the boolean equals(Object) method. E.g.:
L1Obj.equals(L2Obj);
You could overload that method so that you can implement the equality operations that you want.