We Keep Coding

Why does super.hashCode give different results on objects from the same Class?

I have a class DebugTo where if I have two equal instances el1, el2 a HashSet of el1 will not regard el2 as contained.
import java.util.Objects;
public class DebugTo {
public String foo;
public DebugTo(String foo) {
this.foo = foo;
}
#Override
public int hashCode() {
System.out.println(super.hashCode());
return Objects.hash(super.hashCode(), foo);
}
#Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
DebugTo that = (DebugTo) o;
return Objects.equals(foo, that.foo);
}
}
var el1 = new DebugTo("a");
var el2 = new DebugTo("a");
System.out.println("Objects.equals(el1, el2): " + Objects.equals(el1, el2));
System.out.println("Objects.equals(el2, el1): " + Objects.equals(el2, el1));
System.out.println("el1.hashCode(): " + el1.hashCode());
System.out.println("el2.hashCode(): " + el2.hashCode());
Objects.equals(el1, el2): true
Objects.equals(el2, el1): true
1205483858
el1.hashCode(): -1284705008
1373949107
el2.hashCode(): -357249585
From my analysis I have gathered that:
HashSet::contains calls hashCode not equals (relying on the Objects.equals(a, b) => a.hashSet() == b.hashSet())
super.hashCode() gives a different value both times.
Why does super.hashCode() give different results for el1 and el2? since they are of the same class, they have the same super class and so I expect super.hashCode() to give the same result for both.
The hashCode method was probably autogenerated by eclipse. If not answered above, why is super.hashCode used wrong here?

Because the default implementations of the equals and hashCode methods (which go hand in hand - you always override both or neither) treat any 2 different instances as not equal to each other. If you want different behaviour, you override equals and hashCode, and do not invoke super.equals / super.hashCode, or there'd be no point.
HashSets work as follows: They use .hashCode() to know which 'bucket' to put the object into, and if 2 objects end up in the same bucket, equals is used only on those very few objects to double check.
In other words, these are the rules:
If a.equals(b), then b.equals(a) must be true.
a.equals(a) must always be true.
If a.equals(b) and b.equals(c), a.equals(c) must be true.
If a.equals(b), a.hashCode() == b.hashCode() must be true.
The reverse of 4 does not hold: If a.hashCode() == b.hashCode(), that doesn't mean a.equals(b), and hashset does not require it.
Therefore, return 1; is a legal implementation of hashCode.
If a class has really bad hashcode spread (such as the idiotic but legal option listed in bullet 6), then the performance of hashset will be very bad. e.g. set.containsKey(k) which ordinarily takes constant time, will take linear time instead if your objects are all not-equal but have the same hashCode. Hence, do try to ensure hashcodes are as different as they can be.
HashSet and HashMap require stable objects, meaning, their behaviour when calling hashCode and equals cannot change over time.
From the above it naturally follows that overriding equals and not hashCode or vice versa is necessarily broken.
Breaking any of the above rules does not, generally, result in a compiler error. It often doesn't even result in an exception. But instead it results in bizarre behaviour with hashsets and hashmaps: You put an k/v pair in the map, and then immediately ask for the value back and you get null back instead of what you put in, or something completely different. Just an example.
NB: One weird effect of all this is that you cannot add equality-affecting state to subclasses, unless you apply a caveat that most classes including all classes in the core libraries don't apply.
Imagine as an example that we invent the notion of a 'coloured' arraylist. You could have a red '["Hello", "World"]' list, and a blue one:
class ColoredArrayList extends ArrayList {
Color color;
public ColoredArrayList(Color c) {
this.color = color;
}
}
You'd probably want an empty red list to not equal an empty blue one. However, that is impossible if you intend to follow the rules. That's because the equals/hashCode impl of ArrayList itself considers any other list equal to itself if it has the same items in the same order. Therefore:
List<String> a = new ArrayList<String>();
ColoredList<String> b = new ColoredList<String>(Color.RED);
a.equals(b); // this is true, and you can't change that!
Therefore, b.equals(a) must also be true (your impl of equals has to say that an empty red list is equal to an empty plain arraylist), and given that an empty arraylist is also equal to an empty blue one, given that a.equals(b) and b.equals(c) implies that a.equals(c), a red empty list has to be equal to a blue empty list.
There is an easy solution for this that brings in new problems, and a hard solution that is objectively better.
The easy solution is to define that you can't be equal to anything except exact instances of yourself, as in, any subclass is insta-disqualified. Imagine ArrayList's equals method returns false if you call it with an instance of a subclass of ArrayList. Then you could make your colored list just fine. But, this isn't necessarily great, for example, you probably want an empty LinkedList and an empty ArrayList to be equal.
The harder solution is to introduce a second method, canEqual, and call it. You override canEqual to return 'if other is instanceof the nearest class in my hierarchy that introduces equality-relevant state'. Thus, your ColoredList should have #Override public boolean canEqual(Object other) { return other instanceof ColoredList; }.
The problem is, all classes need to have that and use it, or it's not going to work, and ArrayList does not have it. And you can't change that.
Project Lombok can generate this for you if you prefer. It's not particularly common; I'd only use it if you really know you need it.

Wondering about the output of the following Java program

I am dealing with some past exam papers and here I am not sure of the output. I think I am not clear about extends and super.
public class Superclass{
public boolean aVariable;
public void aMethod(){
aVariable = true;
}
}
class Subclass extends Superclass {
public boolean aVariable;
public void aMethod() {
aVariable = false;
super.aMethod();
System.out.println(aVariable);
System.out.println(super.aVariable);
}
}
I think that the second output would be true since it would refer to the super class and it is an object. However, I am not sure of the first output. Would it be just a value and print false or it is also an object?

The output will be:
false
true
Because in your Subclass aVariable is false by default (so assignation aVariable = false; is needless). Read more about Primitive Data Types default values.
And in Superclass you initialize aVariable as true by invoking the superclass' method using the keyword super: super.aMethod();. Read more about Accessing Superclass Members.
Take a look on demo.

Since they're both scoped to their own class block, having them with the same name doesn't matter. As it looks like now, you set aVariable to false, the call to the super doesn't change that, except for creating another variable (new reference) with the same name and sets it to true. So the expected output would be
false
true

Output will be:
false
true
super.aMethod() will execute making aVariable = true of SuperClass
aVariable of SubClass will remain false.

I conducted scientific experiment (copy-pasted and ran) and it prints
false
true
More here:
If you overwrite a field in a subclass of a class, the subclass has two fields with the same name(and different type)?

Java .equals() instanceof subclass? Why not call superclass equals instead of making it final?

It is stated in Object's .equals(Object) javadoc:
It is symmetric: for any non-null reference values x and y,
x.equals(y) should return true if and only if y.equals(x) returns
true.
Almost everywhere in example code I see overridden .equals(Object) method which uses instanceof as one of the first tests, for example here: What issues / pitfalls must be considered when overriding equals and hashCode?
public class Person {
private String name;
private int age;
public boolean equals(Object obj) {
if (obj == null)
return false;
if (obj == this)
return true;
if (!(obj instanceof Person))
return false;
...
}
}
Now with class SpecialPerson extends Person having in equals:
if (!(obj instanceof SpecialPerson))
return false;
we con not guarantee that .equals() is symmetric.
It has been discussed for example here: any-reason-to-prefer-getclass-over-instanceof-when-generating-equals
Person a = new Person(), b = new SpecialPerson();
a.equals(b); //sometimes true, since b instanceof Person
b.equals(a); //always false
Maybe I should add in the beginning of SpecialPerson's equals direct call to super?
public boolean equals(Object obj) {
if( !obj instanceof SpecialPerson )
return super.equals(obj);
...
/* more equality tests here */
}

A lot of the examples use instanceof for two reasons: a) it folds the null check and type check into one or b) the example is for Hibernate or some other code-rewriting framework.
The "correct" (as per the JavaDoc) solution is to use this.getClass() == obj.getClass(). This works for Java because classes are singletons and the VM guarantees this. If you're paranoid, you can use this.getClass().equals(obj.getClass()) but the two are really equivalent.
This works most of the time. But sometimes, Java frameworks need to do "clever" things with the byte code. This usually means they create a subtype automatically. Since the subtype should be considered equal to the original type, equals() must be implemented in the "wrong" way but this doesn't matter since at runtime, the subtypes will all follow certain patterns. For example, they will do additional stuff before a setter is being called. This has no effect on the "equalness".
As you noticed, things start to get ugly when you have both cases: You really extend the base types and you mix that with automatic subtype generation. If you do that, you must make sure that you never use non-leaf types.

You are missing something here. I will try to highlight this:
Suppose you have Person person = new Person() and Person personSpecial = new SpecialPerson() then I am sure you would not like these two objects to be equal. So, its really working as required, the equal must return false.
Moreover, symmetry specifies that the equals() method in both the classes must obey it at the same time. If one equals return true and other return false, then I would say the flaw is in the equals overriding.

Your attempt at solving the problem is not correct. Suppose you have 2 subclasss SpecialPerson and BizarrePerson. With this implementation, BizarrePerson instances could be equal to SpecialPerson instances. You generally don't want that.

don't use instanceof. use this.getClass() == obj.getClass() instead. then you are checking for this exact class.
when working with equalsyou should always use the hashCode and override that too!
the hashCode method for Person could look like this:
#Override
public int hashCode()
{
final int prime = 31;
int result = 1;
result = prime * result + age;
result = prime * result + ((name == null) ? 0 : name.hashCode());
return result;
}
and use it like this in your equals method:
if (this.hashCode() != obj.hashCode())
{
return false;
}

A type should not consider itself equal to an object of any other type--even a subtype--unless both objects derive from a common class whose contract specifies how descendants of different types should check for equality.
For example, an abstract class StringyThing could encapsulate strings, and provide methods to do things like convert to a string or extract substrings, but not impose any requirements on the backing format. One possible subtype of StringyThing, for example, might contain an array of StringyThing and encapsulate the value of the concatenation of all those strings. Two instances of StringyThing would be defined as equal if conversion to strings would yield identical results, and comparison between two otherwise-indistinguishable StringyThing instances whose types knew nothing about each other may have to fall back on that, but StringyThing-derived types could include code to optimize various cases. For example, if one StringyThing represents "M repetitions of character ch" and another represents "N repetitions of the string St", and the latter type knows about the first, it could check whether St contains nothing but M/N repetitions of the character ch. Such a check would indicate whether or not the strings are equal, without having to "expand out" either one of them.

LinkedHashSet: hashCode() and equals() match, but contains() doesn't

How is the following possible:
void contains(LinkedHashSet data, Object arg) {
System.out.println(data.getClass()); // java.util.LinkedHashSet
System.out.println(arg.hashCode() == data.iterator().next().hashCode()); // true
System.out.println(arg.equals(data.iterator().next())); // true
System.out.println(new ArrayList(data).contains(arg)); // true
System.out.println(new HashSet(data).contains(arg)); // true
System.out.println(new LinkedHashSet(data).contains(arg)); // true (!)
System.out.println(data.contains(arg)); // false
}
Am I doing something wrong?
Obviously, it doesn't always happen (if you create a trivial set of Objects, you won't reproduce it). But it does always happen in my case with more complicated class of arg.
EDIT: The main reason why I don't define arg here is that's it's fairly big class, with Eclipse-generated hashCode that spans 20 lines and equals twice as long. And I don't think it's relevant - as long as they're equal for the two objects.

When you build your own objects, and plan to use them in a collection you should always override the following methods:
boolean equals(Object o);
int hashCode();
The default implementation of equals checks whether the objects point to the same object, while you'd probably want to redefine it to check the contents.
As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. To respect the rules an hashCode of an object equals to another one should be the same, thus you've also to redefine hashCode.
EDIT: I was expecting a faulty hashCode or equals implementation, but since your answer, you revealed that you're mutating the keys after they are added to an HashSet or HashMap.
When you add an Object to an hash collection, its hashCode is computed and used to map it to a physical location in the Collection.
If some fields used to compute the hashCode are changed, the hashCode itself will change, so the HashSet implementation will become confused. When it tries to get the Object it will look at another physical location, and won't find the Object. The Object will still be present if you enumerate the set though.
For this reason, always make HashMap or HashSet keys Immutable.

Got it. Once you know it, the answer is so obvious you can only blush in embarrassment.
static class MyObj {
String s = "";
#Override
public int hashCode() {
return s.hashCode();
}
#Override
public boolean equals(Object obj) {
return ((MyObj) obj).s.equals(s);
}
}
public static void main(String[] args) {
LinkedHashSet set = new LinkedHashSet();
MyObj obj = new MyObj();
set.add(obj);
obj.s = "a-ha!";
contains(set, obj);
}
That is enough to reliably reproduce it.
Explanation: Thou Shalt Never Mutate Fields Used For hashCode()!

There seems to be something missing from your question. I have made some guesses:
private void testContains() {
LinkedHashSet set = new LinkedHashSet();
String hello = "Hello!";
set.add(hello);
contains(set, hello);
}
void contains(LinkedHashSet data, Object arg) {
System.out.println(data.getClass()); // java.util.LinkedHashSet
System.out.println(arg.hashCode() == data.iterator().next().hashCode()); // true
System.out.println(arg.equals(data.iterator().next())); // true
System.out.println(new ArrayList(data).contains(arg)); // true
System.out.println(new HashSet(data).contains(arg)); // true
System.out.println(new LinkedHashSet(data).contains(arg)); // true (!)
System.out.println(data.contains(arg)); // true (!!)
}
EDITED: To keep track of changing question!
I still get "true" for ALL but the first output. Please be more specific about the type of the "arg" parameter.

java why should equals method input parameter be Object

I'm going through a book on data structures. Currently I'm on graphs, and the below code is for the vertex part of the graph.
class Vertex<E>{
//bunch of methods
public boolean equals(Object o){
//some code
}
}
When I try to implement this equals method my compiler complains about not checking the type of the parameter and just allowing any object to be sent it. It also does seem a bit strange to me why that parameter shouldn't be a Vertex instead of an Object. Is there a reason why the author does this or is this some mistake or antiquated example?

#Override
public boolean equals(Object obj)
{
if (!(obj instanceof Vertex)) return false;
else return // blah blah
}

equals(Object) is the method defined in the root - Object. If you don't match the signature exactly, Object's version will be called when someone checks if two objects are equal. Not what you want.
You've probably seen other methods (like Comparator) where you can use the exact time. That's because those APIs were generic-ified with Java 5. Equals can't be because it is valid to call equals with two separate types. It should return false, but it is valid.

equals is a method inherited from Object, is defined to be flexible enough so that you can take any object and test if it is equal to any other object (as it rightfully should be able to do), so how could it be any other way?
Edit 1
Comment from jhlu87:
so is it not good form to write an equals method that has an input parameter of vertex?
You are welcome to create your own overload to any method, including equals, but doing so without changing the name could risk confusing many who would assume that your equals is the one that inherits from Object. If it were my code and I wanted a more specific equals method, I'd name it slightly different from just "equals" just to avoid confusion.

If your method doesn't take an argument of type Object, it isn't overriding the default version of equals but rather overloading it. When this happens, both versions exist and Java decides which one to use based on the variable type (not the actual object type) of the argument. Thus, this program:
public class Thing {
private int x;
public Thing(int x) {
this.x = x;
}
public boolean equals(Thing that) {
return this.x == that.x;
}
public static void main(String[] args) {
Thing a = new Thing(1);
Thing b = new Thing(1);
Object c = new Thing(1);
System.out.println(a.equals(b));
System.out.println(a.equals(c));
}
}
confusingly prints true for the first comparison (because b is of type Thing) and false for the second (because c is of type Object, even though it happens to contain a Thing).

It's because this method existed before generics, so for backward compatabitity it has to stay this way.
The standard workaround to impose type is:
return obj instanceof MyClass && <some condition>;

It is because the author is overriding equals. Equals is specified in java.lang.Object and is something that all classes inherrits from.
See the javadoc for java.lang.Object