If two objects return same hashCode, doesn't it mean that they are equal? Or we need equals to prevent collisions?
And can I implement equals by comparing hashCodes?
If two objects have the same hashCode then they are NOT necessarily equal. Otherwise you will have discovered the perfect hash function. But the opposite is true - if the objects are equal, then they must have the same hashCode.
hashCode and Equals are different information about objects
Consider the analogy to Persons where hashcode is the Birthday,
in that escenario, you and many other people have the same b-day (same hashcode), all you are not the same person however..
Why does Java need equals() if there is hashCode()?
Java needs equals() because it is the method through which object equality is tested by examining classes, fields, and other conditions the designer considers to be part of an equality test.
The purpose of hashCode() is to provide a hash value primarily for use by hash tables; though it can also be used for other purposes. The value returned is based on an object's fields and hash codes of its composite and/or aggregate objects. The method does not take into account the class or type of object.
The relationship between equals() and hashCode() is an implication.
Two objects that are equal implies that the have the same hash code.
Two objects having the same hash code does not imply that they are equal.
The latter does not hold for several reasons:
There is a chance that two distinct objects may return the same hash code. Keep in mind that a hash value folds information from a large amount of data into a smaller number.
Two objects from different classes with similar fields will most likely use the same type of hash function, and return equal hash values; yet, they are not the same.
hashCode() can be implementation-specific returning different values on different JVMs or JVM target installations.
Within the same JVM, hashCode() can be used as a cheap precursor for equality by testing for a known hash code first and only if the same testing actual equality; provided that the equality test is significantly more expensive than generating a hash code.
And can I implement equals by comparing hashCodes?
No. As mentioned, equal hash codes does not imply equal objects.
The hashCode method as stated in the Oracle Docs is a numeric representation of an object in Java. This hash code has limited possible values (represented by the values which can be stored in an int).
For a more complex class, there is a high possibility that you will find two different objects which have the same hash code value. Also, no one stops you from doing this inside any class.
class Test {
#Override
public int hashCode() {
return 0;
}
}
So, it is not recommended to implement the equals method by comparing hash codes. You should use them for comparison only if you can guarantee that each object has an unique hash code. In most cases, your only certainty is that if two objects are equal using o1.equals(o2) then o1.hashCode() == o2.hashCode().
In the equals method you can define a more complex logic for comparing two objects of the same class.
If two objects return same hashCode, doesn't it mean that they are equal?
No it doesn't mean that.
The javadocs for Object state this:
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. ...
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 hashCodemethod on
each of the two objects must produce distinct integer results. ...
Note the highlighted statement. It plainly says "No" to your question.
There is another way to look at this.
The hashCode returns an int.
There are only 232 distinct values that an int can take.
If a.hashCode() == b.hashCode() implies a.equals(b), then there can be only 232 distinct (i.e. mutually unequal) objects at any given time in a running Java application.
That last point is plainly not true. Indeed, it is demonstrably not true if you have a large enough heap to hold 232 instances of java.lang.Object ... in a 64-bit JVM.
And a third way is to some well-known examples where two different two character strings have the same hashcode.
Given that your assumption is incorrect, the reasoning that follows from it is also incorrect.
Java does need an equals method.
You generally cannot implement equals using just hashCode.
You may be able to use hashCode to implement a faster equals method, but only if calling hashCode twice is faster than comparing two objects. It generally isn't.
hashCodes are equal -> Objects might be equal -> further comparision is required
hashCodes are different -> Object are not equal (if hashCode is implemented right)
That's how equals method are implemented. At first you check if hashCodes are equal. If yes, you need to check class fields to see if it represents the exact same object. If hashCodes are different, you can be sure that objects are not equal.
Sometimes (very often?) you don't!
These answers are not untrue. But they don't tell the whole story.
One example would be where you are creating a load of objects of class SomeClass, and each instance that is created is given a unique ID by incrementing a static variable, nInstanceCount, or some such, in the constructor:
iD = nInstanceCount++;
Your hash function could then be
int hashCode(){
return iD;
}
and your equals could then be
boolean equals( Object obj ){
if( ! ( obj instanceof SomeClass )){
return false;
}
return hashCode() == obj.hashCode();
}
... under such circumstances your idea that "equals is superfluous" is effectively true: if all classes behaved like this, Java 10 (or Java 23) might say, ah, let's just get rid of silly old equals, what's the point? (NB backwards compatibility would then go out the window).
There are two essential points:
you couldn't then create more than MAXINT instances of SomeClass. Or... you could ... if you set up a system for reassigning the IDs of previously destroyed instances. IDs are typically long rather than int ... but this wouldn't work because hashCode() returns int.
none of these objects could then be "equal" to another one, since equality = identity for this particular class, as you have defined it. Often this is desirable. Often it shuts off whole avenues of possibilities...
The necessary implication of your question is, perhaps, what's the use of these two methods which, in a rather annoying way, have to "cooperate"? Frelling, in his/her answer, alluded to the crucial point: hash codes are needed for sorting into "buckets" with classes like HashMap. It's well worth reading up on this: the amount of advanced maths that has gone into designing efficient "bucket" mechanisms for classes like HashMap is quite frightening. After reading up on it you may come to have (like me) a bit of understanding and reverence about how and why you should bother implementing hashCode() with a bit of thought!
Related
With respect to 3 contracts mentioned below:
1) Whenever hashCode() 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.
From this statement, i understand that, In a single execution of an application, if hashCode() is used one or more times on same object it should return same value.
2) 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.
From this statement, i understand that, to perform the equality operation(in broad scope) in your subclass, There are at least four different degrees of equality.
(a) Reference equality(==), comparing the internal address of two reference type objects.
(b) Shallow structural equality: two objects are "equals" if all their fields are ==.
{ For example, two SingleLinkedList whose "size" fields are equal and whose "head" field point to the same SListNode.}
(c) Deep structural equality: two objects are "equals" if all their fields are "equals".
{For example, two SingleLinkedList that represent the same sequence of "items" (though the SListNodes may be different).}
(d) Logical equality. {Two examples:
(a) Two "Set" objects are "equals" if they contain the same elements, even if the underlying lists store the elements in different orders.
(b) The Fractions 1/3 and 2/6 are "equals", even though their numerators and denominators are all different.}
Based on above four categories of equality, second contract will hold good only: if(Say) equals() method returns truth value based on logical_equality between two objects then hashCode() method must also consider logical_equality amidst computation before generating the integer for every new object instead of considering internal address of a new object.
But i have a problem in understanding this third contract.
3) 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.
In second contract, As we are saying that hashCode() method should be accordingly[for ex: considering logical_equality before generating integer] implemented, I feel, It is not true to say that, if two objects are unequal according to equals(Object) then hashCode() method may produce same integer results as mentioned in third contract? As per the argument in second contract, hashCode() must produce distinct integer results. One just writing return 42 in hashCode() is breaking second contract!
please help me understand this point!
It would be impossible for hashCode() to always return different values for unequal objects. For example, there are 2^64 different Long values, but only 2^32 possible int values. Therefore the hashCode() method for Long has to have some repeats. In situations like this you have to try hard to ensure that your hashCode() method distributes values as evenly as possible, and is unlikely to produce repeats for the instances you are most likely to use in practice.
The second condition just says that two equal() instances must return the same hashCode() value, so this program must print true:
Long a = Long.MIN_VALUE;
Long b = Long.MIN_VALUE;
System.out.println(a.hashCode() == b.hashCode()); // a.equals(b), so must print true.
However this program also prints true:
Long c = 0L;
Long d = 4294967297L;
System.out.println(c.hashCode() == d.hashCode()); // prints true even though !c.equals(d)
hashCode() does not have to produce a distinct result. return 0; is a perfectly legal implementation of hashCode() - it ensures that two equal objects will have the same hash code. But it will ensure dismal performance when using HashMaps and HashSets.
It's preferable that hashCode() return values will be distinct (i.e., objects that are not equal should have different hash codes), but it's not required.
The second contract states what happens when equals() returns true. It does not say anything about the case when equals() returns false.
The third contract is just a reminder about that fact. It reminds you that when equals() is false for two objects, there is no connection between their hash codes. They may be same or different, as the implementation happens to make them.
The third point means that you can have many unequal objects with the same hashcode. . For example 2 string objects can have the same hashcode. The second point states that two equal objects must have the same hashcode. . return 5 is a valid hash implementation because it returns the same value for 2 equal objects.
While reviewing a large code base, I've often come across cases like this:
#Override
public int hashCode()
{
return someFieldValue.hashCode();
}
where the programmer, instead of generating their own unique hash code for the class, simply inherits the hash code from a field value. My gut feeling (which might just as well be digestive problems) tells me that this is wrong, but I can't put my finger on it. What problems can arise, if any, with this sort of implementation?
This is fine if you want to hash your object based on a single property.
For example, in a Person class you might have an ID property that uniquely identifies a Person, so the hashCode() of Person can simply be the hash of that ID.
In addition, the hashCode() is related to the implementation of equals. If two objects are equal, they must have the same hashCode (the opposite doesn't have to be true - two non equal objects may still have the same hashCode). Therefore, if equality is determined by a single property (such as a unique ID), the hashCode method must also use only that single property.
This can be seen in the JavaDoc of hashCode :
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 hash tables.
Technically speaking, you can return any consistent number from hashCode, even a constant value. The only requirement the contract places upon you is that equal objects must return the same hash code:
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.
Theoretically, if all objects return, say, zero for their hashCode, the contract is formally satisfied. However, this makes hashCode completely useless.
The real question is whether you should do it or not. The answer depends on how unique is the field the hash code of which you are returning. It is not uncommon to return the hashCode of a unique identifier of an object for the object's hashCode. On the other hand, if a significant percentage of objects have the sane value of someFieldValue, you would be better off using a different strategy for making the hash code of your object.
hashCode() has to go with equals().
If the only property defining equalness is, for example, an ID, you HAVE TO take care that your hash codes are equal when the ID is equal.
The easiest way to accomplish this is by taking the hashCode() of your ID.
This is fine, if you really want to uniquely identify your object by this single property. Here is an article that explains what object identity really is.
As noted in the documentation of Object, your equals() and hashCode() need to incorporate the same properties, be sure to verify that.
So this means that you should ask yourself the question: do I really want the objects to be equal if only this single property is equal?
Finally do take great care when subclassing objects with a custom equals() and hashcode() implementation, if you want to add properties to the identity of the object, you will break the requirement that a.equals(b) == b.equals(a) (to see why this fails thing about this as a being the super class and b being the subclass.
yes you can do it technically, you need a non-primitive somefieldValue for that.
I agree with the statement from this post What issues should be considered when overriding equals and hashCode in Java?
Use the same set of fields that you use to compute equals() to compute hashCode().
But i've some doubts :
Is this absolutely necessary to have same fields ?
If yes, what if I don't use same field ?
Will it affect HashMap performance or HashMap Accuracy ?
The fields don't have to be the same. The requirement is for two objects that are equal, they must have the same hash code. If they have the same hash code, they don't have to be equal. From the javadocs:
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
hash tables.
For example, you could return 1 as your hash code always, and you would obey the hash code contract, no matter what fields you used in your equals method.
Returning 1 all the time would improve the computation time of hashCode, but HashMap's performance would drop since it would have to resort to equals() more often.
Is this absolutely necessary to have same fields ?
Yes, if you don't want any surprises.
If yes, what if I don't use same field ?
You might get different hashCode for objects that are equal, as per equals() method, which is a requirement for the equals and hashCode contract.
For example, suppose you've 3 fields - a, b, c. And you use a and b for equals() method, and all the 3 fields for hashCode() method. So, for 2 objects, if a and b are equals, and c is different, both will be equals with different hashcode.
Will it affect HashMap performance or HashMap Accuracy ?
It's not about performance, but yes your map will not behave as expected.
Fields used in hashcode can be a subset of fields used in equals.
It will still abide by this rule "Whenever a.equals(b), then a.hashCode() must be same as b.hashCode()"
There is a point in general contract of equals method, which says if You has defined equals() method then You should also define hashCode() method. And if o1.equals(o2) then this is must o1.hashCode() == o2.hashCode().
So my question is what if I break this contract? Where can bring fails the situation when o1.equals(o2) but o1.hashCode != o2.hashCode() ?
It will lead to unexpected behavior in hash based data structure for example: HashMap, Read how HashTable works
HashMap/HashTable/HashSet/etc will put your object into one of several buckets based on its hashCode, and then check to see if any other objects already in that bucket are equal.
Because these classes assume the equals/hashCode contract, they won't check for equality against objects in other buckets. After all, any object in another bucket must have a different hashCode, and thus (by the contract) cannot be equal to the object in quesiton. If two objects are equal but have different hash codes, they could end up in different buckets, in which case the HashMap/Table/Set/etc won't have a chance to compare them.
So, you could end up with a Set that contains two objects which are equal -- which it's not supposed to do; or a Map that contains two values for the same one key (since the buckets are by key); or a Map where you can't look up a key (since the lookup checks both the hash code and equality); or any number of similar bugs.
If you break the contract, your objects won't work with hash-based containers (and anything else that uses hashCode() and relies on its contract).
The basic intuition is as follows: to see whether two objects are the same, the container could call hashCode() on both, and compare the results. If the hash codes are different, the container is allowed to short-circuit by assuming that the two objects are not equal.
To give a specific example, if o1.equals(o2) but o1.hashCode() != o2.hashCode(), you'll likely be able to insert both objects into a HashMap (which is meant to store unique objects).
In java Is it possible to get consistent hash code for an object when we are running the application multiple times
Sure. If it is a String for example, then String.hashCode() gives a consistent hashcode each time you run the application.
You only get into trouble if the hashcode incorporates something other than the values of the object's component fields; e.g. an identity hashcode. And of course, this means that the object class needs to override Object.hashcode() at some point, because that method gives you an identity hashcode.
FOLLOW UP
Judging from comments on other answers, the OP still seems to be pursuing the illusory goal of a unique hash function; i.e. some function that will map (for example) any String to a hashcode that is unique for all possible Strings.
Unfortunately this is impossible in the general case, and in this case. Furthermore, it is a simple matter to construct a proof that a String to int hash function that generates unique int values is mathematically impossible. (I won't bore you with the details ... but the basis of the proof is that there are more String values than int values.)
In fact, the only situation where such a hash function is possible is when the set of all possible values of input type has a size that is no greater than the number of possible values of the integer type. There are hash functions that will map a byte, char, short or int to a unique int, but a hash function that maps long values to unique int values is impossible.
It depends on implementation on hashCode() method of Object
It can also be
public int hashCode() {
return 1;
}
No, not for objects in general. Objects with their own hashcode method will probably be consistent across runs.
Implement/override the public int hashCode() method all objects have?
You have to decide what makes the object the same. Usually it is based on the content of one or more fields. In this case, you should make the hashCode based on these fields. (And equals())
However, I would suggest you shouldn't rely on the hashCode being the same between runs of the application. This is highly likely to break when you change code and very hard to fix when it does. e.g. if you add/remove a field which is part of the hashCode or change the way the hashCode is calculated or anything ti depends on, the hashCode will change.
What are you trying to do? This sounds like a problem where a different solution would be better.
Looking in the contract of hashCode:
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.
So it is not guaranteed that the hashCode is euqal between invocations. In reality, there are be quite some hashCode implementations that return the same value across invocations: String and all types used for boxing (like Integer) have a consistent return value for hashCode. Objects that only combine member hashCodes where each member has a consistent return value also feature this consistency. So, in practice it should be rather common to have a hashCode return value that is consistent accross invocations.