In a usage case, I have a HashMap which contains 1 entry and the key is A. I need to call get() with key B many many times. A equals() to B but A and B are not the same object. The Key contains a long array so its equals() is expensive. I am trying to improve the performance for this map checking operation. I know there are proper ways to address the performance issue. However, I am considering a hack which is the most expedient.
The following is from HashMap.java:
public V get(Object key) {
if (key == null)
return getForNullKey();
int hash = hash(key.hashCode());
for (Entry<K,V> e = table[indexFor(hash, table.length)];
e != null;
e = e.next) {
Object k;
if (e.hash == hash && ((k = e.key) == key || key.equals(k)))
return e.value;
}
return null;
}
if i change the if block in the for loop to:
if (e.hash == hash) {
if (e.key == key) {
return e.value;
} else if (e.key.equals(key)) {
e.key = (K) key;
return e.value;
}
}
I think it will help the performance a lot. The first time I call get() with key B, B's equals() will be called. For the rest of times, B will be == to the key in the map thus saves the equals() call.
However, it is not possible to just extend HashMap and override get() since HashMap.field is package protected and Entry.key is final.
Questions:
Will this scheme work?
Copying HashMap.java and its related code just to change one method is not very appealing. What is the best way to implement this hack?
Thanks!
This is a horrible idea. You are mutating an entry's key under the covers.
The solution is to create your own internal "identity hash value", something you can calculate and guarantee is unique for each value. Then use this as a proxy for the expensive comparison in your equals() method.
For example (pseudo-Java):
class ExpensiveEquals
{
private class InxpensiveEqualsIdentity
{
...
public InexpensiveEqualsIdentity(ExpensiveEquals obj) { ... }
public boolean equals() { an inexpensive comparison }
}
private InxpensiveEqualsIdentity identity;
public ExpensiveEquals(...)
{
... fill in the object
this.identity = new InexpensiveEqualsIdentity(this);
}
public int hashCode() { return this.identity.hashCode(); }
public boolean equals(Object o)
{
if (this == o) return true;
if (o == null || !o instanceof this.getClass()) return false;
return (this.identity.equals(((ExpensiveEquals)o).identity));
}
}
Yes, this should work, if equals is implemented properly (symmetric).
Try to hack equals method in the class of your map's keys:
equals(Object obj) {
if (this == obj) return true;
if (obj == null) return false;
if (!(obj instanceof MyClass)) return false;
MyClass other = (MyClass) obj;
if (this.longArray == other.longArray) return true;
if (Arrays.equals(this.longArray, other.longArray)) {
this.longArray = other.longArray;
return true;
}
return false;
}
Since your class is immutable, this trick should be safe. Your should make longArray field non-final, but it won't hurt performance, I promise.
If B is the key you are really interested in, you can just perform the swap from the outside.
V val = map.remove(b);
map.put(b, val);
From then on, reference equality is sufficient for B but you aren't futzing with the internal mechanism.
My simple lazy idea built on top of #JimGarrison's answer:
private long hash0, hash1;
void initHash() {
// Compute a hash using md5 and store it in hash0 and hash1
// The collision probability for two objects is 2**-128, i.e., very small,
// and grows with the square of the number of objects.
// Use SHA-1 if you're scared.
}
void assureHash() {
if (hash0 == 0 && hash1 == 0) initHash();
}
public int hashCode() {
// If both hashes are zero, assume it wasn't computed yet.
assureHash();
return (int) hash0;
}
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof ExpensiveEquals)) return false;
ExpensiveEquals that = (ExpensiveEquals) o;
this.assureHash();
that.assureHash();
return this.hash0 == that.hash0 && this.hash1 == that.hash1;
}
This ensures that all equals invocations but the first will be pretty cheap. The changes that two randomly chosen pairs of longs are equals are negligible even when assuming thousands of objects and taking the birthday paradox into account. With a cryptographic hash function the numbers are as good as random.
If md5 and two longs aren't good enough, use SHA-1 and one additional int (that's what git does).
Related
The criteria is that equals() method where the objects are considered equal if the value of the double variable is within +/- 10 of the other object's value of the double variable.
I'm not sure how to correctly implement hashCode() so that the hashCode would be equal if it satisfies the conditions of the equals() method.
I would really appreciate your input! Thanks!
public class Test
{
private double value;
private boolean isEqualValues (final double valueOne, final double valueTwo)
{
if(valueOne == valueTwo)
{
return true;
}
else if((valueOne - valueTwo <= 10) &&
(valueOne - valueTwo >= -10))
{
return true;
}
else
{
return false;
}
#Override
public boolean equals(final Object o)
{
if (this == o)
{
return true;
}
if (o == null || getClass() != o.getClass())
{
return false;
}
Test test = (Test) o;
if(isEqualValues(test.value, value))
{
return true;
}
else
{
return false;
}
}
//How to implement hashCode()
#Override
public int hashCode()
{
//unsure how to correctly implement hashCode() so that the hashCode would be equal if it
//satisfies the conditions of the equals() method above
}
}
There's no way to consistently implement this, since equals() demands transitivity:
It is transitive: for any non-null reference values x, y, and z, if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.
new Test(1), new Test(9) and new Test(14) would fail that test (assuming a trivial one-argument constructor that assigns its argument to value).
One way to work around that is to not check for the absolute distance, but "categorize" your objects using some formula, for example take the floor of value / 10 and compare that.
This way some "close" values like new Test(9) and new Test(11) would compare as not-equal, but other than that you'd get a similar result to what you described.
private long getEquivalenceGroup() {
return Math.floorDiv((long) value, 10);
}
#Override
public boolean equals(final Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
Test test = (Test) o;
return test.getEquivalenceGroup() == this.getEquivalenceGroup();
}
#Override
public int hashCode()
{
return Long.hashCode(getEquivalenceGroup());
}
As long as getEquivalenceGroup() is implemented in a stable manner this will produce "groups" of slightly different objects that still compare as equal and has a valid hashCode() implementation.
Note: if you want a comparison as described in the question but you don't necessarily need it to be returned by equals() then adding a boolean isClose(Test other) is perfectly fine. The only problem is you are trying to implement the equals method specifically with that semantic.
You can't and you shouldn't.
You should implement a comparator and do such operations using that.
I have a class called Varfoo that stores variables. I have another class called Replacement that uses a hashmap to replace the x into 2. With the forget method, it's meant to forget replacing x with 2.
Varfoo x = new VarFoo("x");
Replacement s = new Replacement();
s.put(new VarFoo("x"), new IntFoo(2));
x.applyReplacement(s);
s.forget(x);
Here's the forget method:
public boolean forget(VarFoo var) {
if (var == null) {
throw new NullPointerException();
} else {
if (replacementMap.containsKey(var)) {
replacementMap.remove(var);
return true;
} else {
return false;
}
}
}
It will result to null because I've removed the key itself, not what I intended. How do I revert it back to how it was?
Equals and hashcodes of Varfoo:
#Override
public boolean equals(Object o) {
if (o == null) return false;
if (!(o instanceof VarFoo))
return false;
if (o == this)
return true;
return name.equals(((VarFoo) o).name);
}
#Override
public int hashCode() {
int prime = 31;
int result = 1;
result = prime * result + ((name == null) ? 0 : name.hashCode());
return result;
}
You can make a new implementation of Map, that contains two (e.g.) HashMaps. In normal mode it forwards every operation to map1. This is the original map. map2 is null.
When you make a savepoint, you assign an empty map to map2. get operations now go first to map2 and then, if not found, to map1. put operations go only to map2. When you call forget, you assign again null to map2.
Of course, you must implement all the other methods of the Map interface. But this should be a simple task. Take care of removes, if needed (maybe you will need a Set of removed keys.
Hint: You can use java.util.AbstractMap as a base for your implementation.
Consider that I would like to create a class which manages a TreeSet of custom objects with two keys: A String instance identifier, and a long order identifier. The Long value would be used to determine the order of the elements in the list, while the string would be used to determine if two of these elements are duplicates. To clarify, here is what the methods would look like for the custom object
#Override
public boolean equals(Object o) {
if(o == null)
return false;
if(!(o instanceof CustomObject))
return false;
//Determined by string_id values ONLY. Used in TreeSet implementation
CustomObject obj = (CustomObject) o;
return obj.string_id.equals(this.string_id);
}
#Override
public int compareTo(Object another) {
if(another == null || !(another instanceof CustomObject))
return -1;
CustomObject other = (CustomObject) another;
if(other.long_id > this.long_id)
return -1;
else if(other.long_id < this.long_id)
return 1;
else
return 0;
}
Can my TreeSet function this way? I ask, because while testing this, I've found that my implementation only keeps the latest entry, and discards the rest. I'm looking to find out if this is simply an error with my implementation, or if I'm not properly using the TreeSet class and need to refactor my approach.
I have a class
MyData
and its object
myData
In that Class MyData .. there are multiple fields
like
int id
String name
String desc
etc ..
Now i have two objects of this class ..
Is it possible to check that if the data of these two object are all the same , Like both objects have the same Id ,same Name ,same Desc ... Without checking each and every field of this object ..(i.e without checking the id,name,desc of Each object myself) As there are dozens of fields of this object .
I am using JAVA with GWT
Some implementation i came across.. Not sure if this is some thing possible .valid
private static String oldSequence = "";
boolean changed(TestSequence sequence) {
String newSequence = serializeToString(sequence);
boolean changed = !newSequence.equals(oldSequence);
oldSequence = newSequence;
return changed;
}
private static byte[] serialize(Object obj) throws IOException {
ByteArrayOutputStream b = new ByteArrayOutputStream();
ObjectOutputStream o = new ObjectOutputStream(b);
o.writeObject(obj);
return b.toByteArray();
}
private static String serializeToString(Object obj) {
try {
return new String(serialize(obj));
} catch (Exception ex) {
return "" + ex;
}
}
Thanks
You should override hashCode() and equals() method. you can generate these from IDE.
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof MyData)) return false;
MyData myData = (MyData) o;
if (id != myData.id) return false;
if (!desc.equals(myData.desc)) return false;
if (!name.equals(myData.name)) return false;
return true;
}
#Override
public int hashCode() {
int result = id;
result = 31 * result + name.hashCode();
result = 31 * result + desc.hashCode();
return result;
}
Now you can compare the objects. That's it.
Conventional way is to override equals and hashCode methods. Java standard libraries, for instance Map s, List s, Set s use the equals and hashCode functions for equality testing. The code below also null-safe;
Here is the code for your case;
public class MyData {
int id;
String name;
String desc;
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
MyData myData = (MyData) o;
if (id != myData.id) return false;
if (desc != null ? !desc.equals(myData.desc) : myData.desc != null) return false;
if (name != null ? !name.equals(myData.name) : myData.name != null) return false;
return true;
}
#Override
public int hashCode() {
int result = id;
result = 31 * result + (name != null ? name.hashCode() : 0);
result = 31 * result + (desc != null ? desc.hashCode() : 0);
return result;
}
}
and you can test the equality by;
....
Mydata d1 = new...
Mydata d2 = new...
boolean areTheyEqual = d1.equals(d2);
However if you are not allowed to make a compare field by field then you can use byte arrays, there is no need to convert them to strings.
.....
public boolean equals(Object other){
if (this == other) return true;
if (other == null || getClass() != other.getClass()) return false;
byte[] bytesThis = serialize(this);
byte[] bytesOther = serialize(other);
if(bytesOther.length != bytesThis.length) return false;
return Arrays.equals(bytesThis, bytesOther);
}
public static byte[] serialize(Object obj) throws IOException {
ByteArrayOutputStream b = new ByteArrayOutputStream();
ObjectOutputStream o = new ObjectOutputStream(b);
o.writeObject(obj);
return b.toByteArray();
}
...
GWT doesn't make a difference to your requirement.
There is no direct way.
You have to define your equality to check weather they are equal or not. That is overriding equals() method.
#Override
public boolean equals(Object obj) { ...
Before doing:Right way to implement equals contract
Like everyone else is saying, you should override the equals() and hashCode() methods.
Note that you don't have to do this manually. In Eclipse you can simply click on Source/generate hashCode() and equals() and it will do the work for you. I am sure other IDEs have similar feature as well.
If you don't want to add any more code when you add a new field, you can try iterating over fields.
You said "Without checking each and every field of this object ..(i.e without checking the id,name,desc of Each object myself) ", I couldn't figure out whether you don't want to check for each field for equality, or don't want to WRITE a check for each field for equality. I assumed the latter since you tried to add an equality comparison method by using bytewise checks.
Anyways, the code to check each field follows. You can copy/paste to any object. If, in the future, you want some fields to be checked and some not, you can use annotations.
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
MyData myData = (MyData) o;
Field[] fields = this.getClass().getDeclaredFields();
for(Field field:fields){
Object o1 = null;
Object o2 = null;
try {
o1 = field.get(this);
o2 = field.get(o);
} catch (IllegalAccessException e) {
return false;
}
if(o1 == null && o2 != null) return false;
if(o2 == null && o1 != null) return false;
if(o2 == null && o1 == null) continue;
if(!o2.equals(o1)) return false;
}
return true;
}
No.
You have to override the equals() method and compare the objects in that.
Override the equals method of the object in MyData and check the fields independently.
Serialize your objects and compare the results!
You just should be wise in selection of your serialization method.
Override hashCode() and equals() methods
hashCode()
This method provides the has code of an object.
Basically the default implementation of hashCode() provided by Object is derived by mapping the memory address to an integer value. If look into the source of Object class , you will find the following code for the hashCode.
public native int hashCode();
It indicates that hashCode is the native implementation which provides the memory address to a certain extent. However it is possible to override the hashCode method in your implementation class.
equals()
This method is used to make equal comparison between two objects. There are two types of comparisons in Java. One is using “= =” operator and another is “equals()”. I hope that you know the difference between this two. More specifically the .equals() refers to equivalence relations. So in broad sense you say that two objects are equivalent they satisfy the equals() condition.
List<Pair<String, String> > lp = new ArrayList<Pair<String, String> >();
lp.add(new Pair("1", "2"));
How should I check if the list lp contains 1 and 2 i.e the Pair ("1", "2").
Your Pair class needs to implement equals() and hashCode() and you're all set. List.contains() is implemented in terms of the type's equals() method. See the API for List.contains(). (Edited a bit to address comments from #maaartinus, whose answer you should read b/c the observations are solid, and it's a bit ridiculous for me to fold them in here. As maaartinus points out, a best-practice here would be to avoid error-prone manual definitions for equals and hashcode, and instead build on Guava's helper functions for nullable equals and hashCode for n objects).
final class Pair<T> {
final T left;
final T right;
public Pair(T left, T right)
{
if (left == null || right == null) {
throw new IllegalArgumentException("left and right must be non-null!");
}
this.left = left;
this.right = right;
}
public boolean equals(Object o)
{
// see #maaartinus answer
if (! (o instanceof Pair)) { return false; }
Pair p = (Pair)o;
return left.equals(p.left) && right.equals(p.right);
}
public int hashCode()
{
return 7 * left.hashCode() + 13 * right.hashCode();
}
}
With suitable equals(), you can now do:
lp.add(new Pair("1", "2"));
assert lp.contains(new Pair("1","2"));
Responding to the comments below, perhaps it would be good to include a good reference for "Why do I need to implement hashCode()?"
JavaPractices.com — Implementing equals() — "if you override equals, you must override hashCode"
Object.equals() contract as defined in the API documentation
StackOverflow answer
The implementation in the answer by andersoj
return left != null && right != null && left.equals(p.left) && right.equals(p.right);
is wrong: The null tests clearly suggest that null is a legal value for left and right. So there are at least two problems there:
new Pair(null, null).hashCode() throws NPE
new Pair(null, null) does NOT equal to itself!
Have a look at Guava class Objects for a correct implementation. Use it or write a static helper methods like
public static boolean equal(Object a, Object b) {
return a==b || a!=null && a.equals(b);
}
public static int hashCode(Object a) {
return a==null ? 0 : a.hashCode();
}
and always use them.
Never ever write equals containing a null test.
It's to easy to blow it, and nobody noticed it. Using the Helper, it's trivial to get it right:
public boolean equals(Object o) {
if (!(o instanceof Pair)) return false;
Pair p = (Pair) o;
return Helper.equals(left, p.left) && Helper.equals(right, p.right);
}
public int hashCode() {
return 7 * Helper.hashCode(left) + 13 * Helper.hashCode(right);
}
Of course, forbidding nulls in the constructor is an option, too.