class Hash {
int a;
Hash(int h){
a=h;
}
public boolean equals(Object o) {
Boolean h=super.equals(o);
System.out.println("Inside equals ");
return h;
}
public int hashCode() {
System.out.println("Inside Hash");
return 2;
}
}
public class Eq {
public static void main(String...r) {
HashMap<Hash,Integer> map=new HashMap<Hash,Integer>();
Hash j=new Hash(2);
map.put(j,1);
map.put(j,2);
System.out.println(map.size());
}
}
output was
inside hash
inside hash
1
Since it returns the same hashcode , the second time an object is added in hashmap it must use the equals method but it doesnt call . So wats the problem here?
The HashMap is testing with == before .equals, and since you are putting the same object twice, the first test passes. Try with:
Hash j=new Hash(2);
Hash k=new Hash(2);
map.put(j,1);
map.put(k,2);
The equality check is done by HashMap in three steps:
hash code is different => unequal
objects are identical (==) => equal
equal method gives true => equal
The second step prevents calling equals since identical objects are always assumed equal.
From the doc
put(): Associates the specified value with the specified key in this map. If the map previously contained a mapping for this key, the old value is replaced.
Related
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.
I was reviewing one of Oracle’s Java Certification Practice Exams when I came across the follow question:
Given:
class MyKeys {
Integer key;
MyKeys(Integer k) {
key = k;
}
public boolean equals(Object o) {
return ((MyKeys) o).key == this.key;
}
}
And this code snippet:
Map m = new HashMap();
MyKeys m1 = new MyKeys(1);
MyKeys m2 = new MyKeys(2);
MyKeys m3 = new MyKeys(1);
MyKeys m4 = new MyKeys(new Integer(2));
m.put(m1, "car");
m.put(m2, "boat");
m.put(m3, "plane");
m.put(m4, "bus");
System.out.print(m.size());
What is the result?
A) 2
B) 3
C) 4
D) Compilation fails
My guess was B because m1 and m3 are equal due to their key references being the same. To my surprise, the answer is actually C. Does put() do something that I am missing? Why wouldn’t "plane" replace "car"? Thank you!
With given definition of class i.e
class MyKeys {
Integer key;
MyKeys(Integer k) {
key = k;
}
public boolean equals(Object o) {
return ((MyKeys) o).key == this.key;
}
}
It will result ans = 4, it has only equal method, if you add definition of hashcode then it will result ans=3
class MyKeys {
Integer key;
MyKeys(Integer k) {
this.key = k;
}
#Override
public boolean equals(Object o) {
return ((MyKeys) o).key == this.key;
}
#Override
public int hashCode(){
return key*key;
}
}
Contract of equal and hashcode:
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. If you only override equals() and not hashCode() your class violates this contract.
The problem you will have is with collections where unicity of elements is calculated according to both .equals() and .hashCode(), for instance keys in a HashMap.
If you have two objects which are .equals(), but have different hash codes, you lose!
If we keep this simple, since this is for a Java Certification.
Notice that MyKeys doesn't override hashCode, you know there will be something about it. And I usually try to remember only one thing about Object.hashCode
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 JavaTM programming language.)
Or in short, every instance will have a distinct hashcode. Meaning that with this code, every new MyKeys will add a new pair in the map.
In reality, this is a bit more complex since the method still return an integer, so the risk of collision is still present (integer doesn't provide infinite amount of values). You can see a bit more about this here.
This explain why the answer is that the map will have a size of 4. Each key inserted is a different instance.
As answered by others, the ans will be 4, reason being not overriding hashcode method.
For a more clear reason, whenever an object is added in hash map, the hashcode of the key is generated, which decides the location of the entry set. The 2 objects m1 and m3 will have different hash codes as the hashcode method is not overridden (usual hashcode behaviour). Different hash code will not create any collision and a new entry is made.
On the contrary, the equals methods is called only after the hashcode method produces the same result, i.e., same hash code.
In case of m2 and m4 also, the 2 objects have different hash codes, hence 2 different entries, with no calling done to the equals method.
Hence, in cases of hashing, it is necessary to overload hashcode method, along with equals.
It will be more clear when we see the implementation of put method of HashMap.
// here hash(key) method is call to calculate hash.
// and in putVal() method use int hash to find right bucket in map for the object.
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
In your code you #Override only equals method.
class MyKeys {
Integer key;
MyKeys(Integer k) {
key = k;
}
public boolean equals(Object o) {
return ((MyKeys) o).key == this.key;
}
}
To achieve output you need to override both hashCode() and equals() method.
I have a program to search the key to print the values from a hashmap. But my inputs to the Key and Values are objects that are user defined.Now when I'm equating input key with key1 why are the hashcodes of the Objects key and key1 in the program appearing different, although the return type is same,ie. NameInit, where the hashcodes of the String str="abc" and abc are returned equal? How to check the equality of key and key1 in the program? I tried Objects.equals(key,key1) after type-casting to Object class, but still did not work.I have seen questions of similar kind like in [this question][1] that discusses about the hashcode equality, but then again how to do the equality of these objects as in my example. Kindly help.
NameInit Class
public class NameInit {
String name;
public NameInit(String name)
{
this.name = name;
}
#Override
public String toString(){
return name;
}
}
PlaceAndAddInit
public class PlaceAndAddInit {
String place;
int value;
public PlaceAndAddInit(String place,int val) {
this.place = place;
this.value= val;
}
#Override
public String toString(){
return place+" "+value;
}
}
Main Class
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
HashMap h = new HashMap();
System.out.println("Limit: ");
int limit = scan.nextInt();
for(int i=0;i<limit;i++)
{
h.put(new NameInit(scan.next()), new PlaceAndAddInit(scan.next(),
scan.nextInt()));
}
System.out.println("Enter a key to search the value: ");//
NameInit key= new NameInit(scan.next());//asks for a input from the user to fetch the values
Set s = h.entrySet();
Iterator<NameInit> itr = s.iterator();
while(itr.hasNext())
{
Map.Entry<NameInit,PlaceAndAddInit> me = (Map.Entry) itr.next();
NameInit key1 =me.getKey();
if(key.equals(key1)){// this never happens with this code as key and key1 holds different hashcodes. So how do I achieve the equality.
System.out.println(me.getValue());
}
}
}
}
Edit: I tried to obtain equality by equals method to which I discovered that hashcodes of key1 and key are different. Understanding the reason behind this is the purpose of my question.
You're not overriding hashCode() so the default is used. In the default implementation, key and key1 will have different hashCode values and they will not be equal even if you think they should be. So the solution is to override the hashCode and equals method if you want to be able to compare those objects.
To answer your question:
Edit: I tried to obtain equality by equals method to which I
discovered that hashcodes of key1 and key are different. Understanding
the reason behind this is the purpose of my question.
If you don't provide overrides to equals and hashCode, they will get inherited from Object. And here's how they look for Object:
public boolean equals(Object obj) {
return (this == obj);
}
Therefore, 2 objects will only be equal when they are == which means that they point to precisely the same memory location. In your example, it is not the case. hashCode is native so can't show you the source code.
Here's more to read:
Google search about hashCode and equals
Objects.equals is implemented like this:
return (a == b) || (a != null && a.equals(b));
You see, it is basically calling a's equals method, not the hashcode method. No matter how hashcode is implemented Objects.equals returns false when a.equals(b) returns false. It has nothing to do with the hashcode method.
So to fix this, simply override the equals method. This is a simple implementation:
#Override
public boolean equals(Object obj) {
return this.hashcode() == obj.hashcode();
}
Also, if you want to find the value of a key in the hash map, call the get method on the hash map and it will do it for you in O(1) time. No need for such an inefficient approach with O(n) time.
Since duplicate key overrides the previous key and its corresponding value in hashmap. But if i call get() method and provide previous key as an argument, it returns the overriden value. How it is possible since that key is overriden by new key so it should throw an exception.
//This class is used as a key
Class MapKey
{
public boolean equlas( Object o)
{
return true;
}
}
//This is test class
class MapTest
{
public static void main(String a[])
{
Map m=new Hashmap<Mapkey, String>();
MapKey mk1=new MapKey();
m.put(mk1,"one");
MapKey mk2=new MapKey();
m.put(mk2,"two");
MapKey mk3=new MapKey();
m.put(mk3,"three");
System.out.println(m.get(mk1));
System.out.println(m.get(mk2));
}
}
output: three
three
Since keys are equal so it should get overriden by last key object mk3.
so how is it possible to retrieve value with first or second key object?
Thanks in advance
This is because you have not overrided hashcode method. In equals method you have declared all the objects are equal. But due to default hashcode implementation all equal objects are returning different hashcodes. As hashcodes are used to save values in a hashmap, you are able to get values for old keys mk1 and mk2.
The equals method isn't the most important for HashMap. As the name suggests, the most important method for inserting values into the HashMap is hashCode, not equals. A value is only overwriten in HashMap, if a key k in the map fullfills the following condition: k.hashCode() is equal to the hashCode of the key for inserting an item and the key equals k.
The code you post doesn't come up with
three
three
even if I fix it to clean up the compile errors:
import java.util.*;
class MapKey
{
public boolean equlas( Object o)
{
return true;
}
}
class MapTest
{
public static void main(String a[])
{
Map<MapKey, String> m=new HashMap<MapKey, String>();
MapKey mk1=new MapKey();
m.put(mk1,"one");
MapKey mk2=new MapKey();
m.put(mk2,"two");
MapKey mk3=new MapKey();
m.put(mk3,"three");
System.out.println(m.get(mk1));
System.out.println(m.get(mk2));
}
}
This prints
one
two
Even if I fix the MapKey to override equals:
class MapKey
{
public boolean equlas( Object o)
{
return true;
}
public boolean equals( Object o)
{
return true;
}
}
this prints the same output.
If I implement hashCode:
class MapKey
{
public int hashCode()
{
return 1;
}
public boolean equals( Object o)
{
return true;
}
}
then it will print out
three
three
The default implementations of hashCode and equals are based on the object reference, references to different objects will not be equal. Maps use hashCode to decide which bucket to store an object in, and use equals to differentiate between different objects in the same bucket.
Here you made three different instances so their hashCodes will be different (as long as there isn't a collision, in which case the equals method will get called to decide if the two objects are the same). In this case there isn't a collision, and it's not until hashCode gets overridden that the different mapKey instances are treated as equivalent.
I have a problem when retrieving values from a hashmap. The hashmap is declared as follows:
HashMap<TRpair,A> aTable = new HashMap<TRpair,A>();
I then put 112 values into the map as follows:
aTable.put(new TRpair(new T(<value>),new Integer(<value>)),new Ai());
where Ai is any one of 4 subclasses that extend A.
I then proceed to check what values are in the map, as follows:
int i = 0;
for (Map.Entry<TRpair,A> entry : aTable.entrySet()) {
System.out.println(entry.getKey().toString() + " " + entry.getValue().toString());
System.out.println(entry.getKey().equals(new TRpair(new T("!"),new Integer(10))));
i++;
}
i holds the value 112 at the end, as one would expect and the equality test prints true for exactly one entry, as expected.
However, when I do
System.out.println(aTable.get(new TRpair(new T("!"), new Integer(10))));
null is output, despite the above code snippet confirming that there is indeed one entry in the map with exactly this key.
If it helps, the class TRpair is declared as follows:
public class TRpair {
private final T t;
private final Integer r;
protected TRpair(Integer r1, T t1) {
terminal = t1;
row = r1;
}
protected TRpair(T t1, Integer r1) {
t = t1;
r = r1;
}
#Override
public boolean equals(Object o) {
TRpair p = (TRpair)o;
return (p.t.equals(t)) && (p.r.equals(r));
}
#Override
public String toString() {
StringBuilder sbldr = new StringBuilder();
sbldr.append("(");
sbldr.append(t.toString());
sbldr.append(",");
sbldr.append(r.toString());
sbldr.append(")");
return sbldr.toString();
}
}
the equals() and toString() methods in each of the Ai (extending A) and in the T class are overridden similarly and appear to behave as expected.
Why is the value output from the hashmap aTable null, when previously it has been confirmed that the value for the corresponding key is indeed in the map?
With many thanks,
Froskoy.
The keys/elements for a Hash collection but override hashCode() if euqals is overridden.
You could use.
public int hashCode() {
return t.hashCode() * 31 ^ r.hashCode();
}
BTW: It appears from your code that Integer r cannot be null in which case using int r makes more sense.
From Object.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.
IIRC hashmap looks up by hashCode() and not by equality, and since you did not implemented hashcode you use default implementation which is consistent with object pointer equality -
you need to implement proper hashcode function which takes into account "T" parameter as well as integer (or not)
It is good practice that hashCode() and equals() are consistent, but not structly necessary if you know what you are doing.