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How to implement a compareTo() method when consistent with Equal and hashcode

I have a class Product, which three variables:
class Product implements Comparable<Product>{
private Type type; // Type is an enum
Set<Attribute> attributes; // Attribute is a regular class
ProductName name; // ProductName is another enum
}
I used Eclipse to automatically generate the equal() and hashcode() methods:
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((attributes == null) ? 0 : attributes.hashCode());
result = prime * result + ((type == null) ? 0 : type.hashCode());
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Product other = (Product) obj;
if (attributes == null) {
if (other.attributes != null)
return false;
} else if (!attributes.equals(other.attributes))
return false;
if (type != other.type)
return false;
return true;
}
Now in my application I need to sort a Set of Product, so I need to implement the Comparable interface and compareTo method:
#Override
public int compareTo(Product other){
int diff = type.hashCode() - other.getType().hashCode();
if (diff > 0) {
return 1;
} else if (diff < 0) {
return -1;
}
diff = attributes.hashCode() - other.getAttributes().hashCode();
if (diff > 0) {
return 1;
} else if (diff < 0) {
return -1;
}
return 0;
}
Does this implementation make sense? What about if I just want to sort the product based on the String values of "type" and "attributes" values. So how to implement this?
Edit:
The reason I want to sort a Set of is because I have Junit test which asserts on the string values of a HashSet. My goal is to maintain the same order of output as I sort the set. otherwise, even if the Set's values are the same, the assertion will fail due to random output of a set.
Edit2:
Through the discussion, it's clear that to assert the equality of String values of a HashSet isn't good in unit tests. For my situation I currently write a sort() function to sort the HashSet String values in natural ordering, so it can consistently output the same String value for my unit tests and that suffice for now. Thanks all.

Looks like from all the comments in here you dont need to use Comparator at all. Because:
1) You are using HashSet that does not work with Comparator. It is not ordered.
2) You just need to make sure that two HashSets containing Products are equal. It means they are same size and contain the same set of Products.
Since you already added hashCode and equals methods to Product all you need to do is call equals method on those HashSets.
HashSet<Product> set1 = ...
HashSet<Product> set2 = ...
assertTrue( set1.equals(set2) );

This implementation does not seem to be consistent. You have no control over how the hash codes look like. If you have obj1 < obj2 according to compareTo in the first try, the next time you start your JVM it could be the other way around obj1 > obj2.
The only thing that you really know is that if diff == 0 then the objects are considered to be equal. However you can also just use the equals method for that check.
It is now up to you how you define when obj1 < obj2 or obj1 > obj2. Just make sure that it is consistent.
By the way, you know that the current implementation does not include ProductName name in the equals check? Dont know if that is intended thus the remark.
The question is, what do you know about that attributes? Maybe they implement Comparable (for example if they are Numbers), then you can order according to their compareTo method. If you totally know nothing about the objects, it will be hard to build up a consistent ordering.
If you just want them to be ordered consistently but the ordering itself does not play any role, you could just give them ids at creation time and sort by them. At this point you could indeed use the hashcodes if it does not matter that it can change between JVM calls, but only then.

Is there a way to create an AND/OR/ETC. Boolean statement that is comparing with the same value without redefining the value? [duplicate]

This question already has answers here:
Best way to format multiple 'or' conditions in an if statement
(8 answers)
Closed 1 year ago.
Basically, what I want to do is check two integers against a given value, therefore, classically what you would do is something like this:
//just to get some values to check
int a, b;
a = (int)(Math.random()*5);
b = (int)(Math.random()*5);
//the actual thing in question
if(a == 0 || b == 0)
{
//Then do something
}
But is there a more concise format to do this? Possibly similar to this (which returns a bad operand type):
//just to get some values to check
int a, b;
a = (int)(Math.random()*5);
b = (int)(Math.random()*5);
//the actual thing in question
if((a||b) == 0)
{
//Then do something
}

You can do the following in plain java
Arrays.asList(a, b, c, d).contains(x);

Unfortunately there is no such construct in Java.
It this kind of comparison is frequent in your code, you can implement a small function that will perform the check for you:
public boolean oneOfEquals(int a, int b, int expected) {
return (a == expected) || (b == expected);
}
Then you could use it like this:
if(oneOfEquals(a, b, 0)) {
// ...
}
If you don't want to restrict yourselft to integers, you can make the above function generic:
public <T> boolean oneOfEquals(T a, T b, T expected) {
return a.equals(expected) || b.equals(expected);
}
Note that in this case Java runtime will perform automatic boxing and unboxing for primitive types (like int), which is a performance loss.

As referenced from this answer:
In Java 8+, you might use a Stream and anyMatch. Something like
if (Stream.of(b, c, d).anyMatch(x -> x.equals(a))) {
// ... do something ...
}
Note that this has the chance of running slower than the other if checks, due to the overhead of wrapping these elements into a stream to begin with.

I think that a bit-wise OR:
if ((a | b) == 0) . . .
would work if you want to check specifically for 0. I'm not sure if this saves any execution time. More to the point, it makes for cryptic code, which will make the future maintainer of this code curse you (even if its yourself). I recommend sticking with the more-typing option.
Bah. I misread OP's original logic.
Another go...
If you want to test whether any one of many variables is equal to an expected value, a generic function might work:
public <T> boolean exists(T target, T... values) {
for (T value : values) {
if (target == null) {
if (value == null) {
return true;
}
} else if (target.equals(value)) {
return true;
}
}
return false;
}
This will work for any number of objects of one type. Primitives will be autoboxed so it will work with them as well. Your original code will be something like:
if (test(0, a, b)) {
// do something
}
(A better method name would be desperately needed to even consider whether this an improvement over what you have now. Even if the test expands to 3 or 4 variables, I question the need for this kind of thing.) Note that this also works with arrays:
int[] values = { . . . };
if (test(0, values)) { . . .
and it can be used to test whether an array (or any of a collection of variables) is null.

if(a == 0 || b == 0)
{
//Then do something
}
Why not keep it readable? What is not concise about this? On the other hand,
a = (int)(Math.random()*5);
involves an unnecessary cast. Why not just use Random and invoke nextInt()?

For this example, you can do
if (a * b == 0)
or for more variables
if (a * b * c * d == 0)
while more concise it may not be as clear. For larger values, you need to cast to a long to avoid an overflow.

You could put the integers in a set and see if it contains the given value. Using Guava:
if(newHashSet(a, b).contains(0)){
// do something
}
But two simple int comparisons are probably easier to understand in this case.

Here's a modification of #buc's answer that can take any number of any arguments:
public <T> boolean oneOfEquals(T expected, T... os) {
for (T o : os) {
if (expected.equals(o)) return true;
}
return false;
}

Even if you have used the bit-wise operation as Ted suggested, the expressions are not equal, since one requires at least one of the variables to be zero and the second requires both of them to be zero.
Regarding your question, there is no such shortcut in Java.

You can try this code:
public static boolean match (Object ref, Object... objects)
{
if (ref == null)
return false;
//
for (Object obj : objects)
if (obj.equals (ref))
return true;
//
return false;
} // match
So if you can check this way:
if (match (reference, "123", "124", "125"))
; // do something

In Java 8 we can achieve the same by using the below method:
private boolean methodName(int variant,int... args){
if(args.length > 0){ return Arrays.stream(args).anyMatch( x -> x == variant); }
return false;
}
The given method will return true if the variant will match any possible input(s). This is used for or condition.
In the same way, if you want to do &&(and) condition then you just need to used other Java 8 methods:
Note: These methods take Predicate as an argument.
anyMatch: return true the moment the first predicate returns true otherwise false.
allMatch: return true if all the predicates return true otherwise false.
noneMatch: return true if none of the predicates return true otherwise false.
Performance Note: This is good when you have enough amount of data to
check as it has some overhead but it works really well when you use
this for enough amount of data. normal way is good for just two
conditions.

There is no special syntax for that. You could make a function for that. Assuming at least Java 1.5:
public <T> boolean eitherOneEquals(T o1, T o2, T expectedValue) {
return o1.equals(expectedValue) || o2.equals(expectedValue);
}
if(eitherOneEquals(o1, o2, expectedValue)) {
// do something...
}

compareTo() method in java (Student ids)

Learning java and having trouble with the compareTo method. I tried google but it was not much help for what i need.What i need is
// compareTo public int compareTo(Student other)
// is defined in the Comparable Interface
// and should compare the student ID's (they are positive integers).
// Must be able to handle null "other" students. A null student should be
// ordered before any student s so the s.compareTo(null) should be positive.
so basically a compareTo(), in the end this method is going to help me put my students in order based on there student ids lowest to greatest.. I'm at a brick wall and just need some help in the right direction
public int compareTo(StudentIF other) {
// do stuff
return 0;
}

There's a good tutorial about implementing compareTo() here. That said, when learning how to do something in general it's often helpful for me to see how to implement it in my specific use case - so, in this case, I would imagine something like this would suffice:
public int compareTo(StudentIF other) {
if (other == null) {return 1;} //satisfies your null student requirement
return this.studentId > other.studentId ? 1 :
this.studentId < other.studentId ? -1 : 0;
}
compareTo() is expected to return a positive value if the other object is comparitively smaller, a negative value if it's comparitively larger, and 0 if they are equal. Assuming you're familiar with the ternary operator, you'll see that that's what this is doing. If you're not, then the if/else equivalent would be:
public int compareTo(StudentIF other) {
if (other == null) { return 1; } //satisfies your null student requirement
if (this.studentId > other.studentId) return 1;
else if (this.studentId < other.studentId) return -1;
else return 0; //if it's neither smaller nor larger, it must be equal
}

As the compareTo interface required:
a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.
plus your additional requirement of null comparison, we can simply check whether the other param is null or not, and then do a subtraction to compare.
public int compareTo(StudentIF other) {
if (other == null) {
return 1;
}
return this.id - other.id;
}

Populating a TreeMap throws NullPointerException

I am getting a NullPointerException while creating a TreeMap.
Here is my code:
public TreeMap<AccountGroupBean,List<AccountBean>> getAccountsAndGroups() throws SessionExpiredException {
TreeMap<AccountGroupBean,List<AccountBean>> map = new TreeMap<AccountGroupBean,List<AccountBean>>();
List<AccountGroupBean> groups = getAccountGroups();
for(AccountGroupBean group : groups) {
List<AccountBean> accounts = getAccountsByGroupId(group.getId());
System.out.println("PRINT"+ accounts.size());
map.put(group,accounts);
System.out.println("!" +map.get(group).size());
}
return map;
}
The first println prints 44. That it is to say is not null. However, the second println raises the null exception.
Any idea what I am doing wrong?
Solution
AS pointed in the accepted solution. The problem was in my implementation of compareTo.
I used to have:
public int compareTo(AccountGroupBean o) {
return (number > o.getNumber()) ? 1 : -1;
}
Adding a 0 return solved the issue:
public int compareTo(AccountGroupBean o) {
if(number == o.getNumber()) {
return 0;
}
return (number > o.getNumber()) ? 1 : -1;
}

I looks like AccountGroupBean doesn't implement Comparable in a proper way, try to println group.compareTo(group) to check if it prints 0.

This is most likely a problem with how the AccountGroupBean class is implementing equals and hashcode. There are some rules fro implementing equals and hashcode that you should make sure your code complies to. Some of the rules for the equals method include.
Reflexive for any non null value x.equals(x) is always true
Symetric for non null values y.equals(x) must return true if and only if x.equals(y) is true
Transitive for non null values if x.equals(y) is true and y.equals(z) is true then x.equals(z) must also be true
Consistant The equals method should return the same answer during multiple invocations if the objects have not been modfied.
If two objects are equal their hashcoe method should return the same value.

Treeset.contains() problem

So I've been struggling with a problem for a while now, figured I might as well ask for help here.
I'm adding Ticket objects to a TreeSet, Ticket implements Comparable and has overridden equals(), hashCode() and CompareTo() methods. I need to check if an object is already in the TreeSet using contains(). Now after adding 2 elements to the set it all checks out fine, yet after adding a third it gets messed up.
running this little piece of code after adding a third element to the TreeSet, Ticket temp2 is the object I'm checking for(verkoopLijst).
Ticket temp2 = new Ticket(boeking, TicketType.STANDAARD, 1,1);
System.out.println(verkoop.getVerkoopLijst().first().hashCode());
System.out.println(temp2.hashCode());
System.out.println(verkoop.getVerkoopLijst().first().equals(temp2));
System.out.println(verkoop.getVerkoopLijst().first().compareTo(temp2));
System.out.println(verkoop.getVerkoopLijst().contains(temp2));
returns this:
22106622
22106622
true
0
false
Now my question would be how this is even possible?
Edit:
public class Ticket implements Comparable{
private int rijNr, stoelNr;
private TicketType ticketType;
private Boeking boeking;
public Ticket(Boeking boeking, TicketType ticketType, int rijNr, int stoelNr){
//setters
}
#Override
public int hashCode(){
return boeking.getBoekingDatum().hashCode();
}
#Override
#SuppressWarnings("EqualsWhichDoesntCheckParameterClass")
public boolean equals(Object o){
Ticket t = (Ticket) o;
if(this.boeking.equals(t.getBoeking())
&&
this.rijNr == t.getRijNr() && this.stoelNr == t.getStoelNr()
&&
this.ticketType.equals(t.getTicketType()))
{
return true;
}
else return false;
}
/*I adjusted compareTo this way because I need to make sure there are no duplicate Tickets in my treeset. Treeset seems to call CompareTo() to check for equality before adding an object to the set, instead of equals().
*/
#Override
public int compareTo(Object o) {
int output = 0;
if (boeking.compareTo(((Ticket) o).getBoeking())==0)
{
if(this.equals(o))
{
return output;
}
else return 1;
}
else output = boeking.compareTo(((Ticket) o).getBoeking());
return output;
}
//Getters & Setters

On compareTo contract
The problem is in your compareTo. Here's an excerpt from the documentation:
Implementor must ensure sgn(x.compareTo(y)) == -sgn(y.compareTo(x)) for all x and y.
Your original code is reproduced here for reference:
// original compareTo implementation with bug marked
#Override
public int compareTo(Object o) {
int output = 0;
if (boeking.compareTo(((Ticket) o).getBoeking())==0)
{
if(this.equals(o))
{
return output;
}
else return 1; // BUG!!!! See explanation below!
}
else output = boeking.compareTo(((Ticket) o).getBoeking());
return output;
}
Why is the return 1; a bug? Consider the following scenario:
Given Ticket t1, t2
Given t1.boeking.compareTo(t2.boeking) == 0
Given t1.equals(t2) return false
Now we have both of the following:
t1.compareTo(t2) returns 1
t2.compareTo(t1) returns 1
That last consequence is a violation of the compareTo contract.
Fixing the problem
First and foremost, you should have taken advantage of the fact that Comparable<T> is a parameterizable generic type. That is, instead of:
// original declaration; uses raw type!
public class Ticket implements Comparable
it'd be much more appropriate to instead declare something like this:
// improved declaration! uses parameterized Comparable<T>
public class Ticket implements Comparable<Ticket>
Now we can write our compareTo(Ticket) (no longer compareTo(Object)). There are many ways to rewrite this, but here's a rather simplistic one that works:
#Override public int compareTo(Ticket t) {
int v;
v = this.boeking.compareTo(t.boeking);
if (v != 0) return v;
v = compareInt(this.rijNr, t.rijNr);
if (v != 0) return v;
v = compareInt(this.stoelNr, t.stoelNr);
if (v != 0) return v;
v = compareInt(this.ticketType, t.ticketType);
if (v != 0) return v;
return 0;
}
private static int compareInt(int i1, int i2) {
if (i1 < i2) {
return -1;
} else if (i1 > i2) {
return +1;
} else {
return 0;
}
}
Now we can also define equals(Object) in terms of compareTo(Ticket) instead of the other way around:
#Override public boolean equals(Object o) {
return (o instanceof Ticket) && (this.compareTo((Ticket) o) == 0);
}
Note the structure of the compareTo: it has multiple return statements, but in fact, the flow of logic is quite readable. Note also how the priority of the sorting criteria is explicit, and easily reorderable should you have different priorities in mind.
Related questions
What is a raw type and why shouldn't we use it?
How to sort an array or ArrayList ASC first by x and then by y?
Should a function have only one return statement?

This could happen if your compareTo method isn't consistent. I.e. if a.compareTo(b) > 0, then b.compareTo(a) must be < 0. And if a.compareTo(b) > 0 and b.compareTo(c) > 0, then a.compareTo(c) must be > 0. If those aren't true, TreeSet can get all confused.

Firstly, if you are using a TreeSet, the actual behavior of your hashCode methods won't affect the results. TreeSet does not rely on hashing.
Really we need to see more code; e.g. the actual implementations of the equals and compareTo methods, and the code that instantiates the TreeSet.
However, if I was to guess, it would be that you have overloaded the equals method by declaring it with the signature boolean equals(Ticket other). That would lead to the behavior that you are seeing. To get the required behavior, you must override the method; e.g.
#Override
public boolean equals(Object other) { ...
(It is a good idea to put in the #Override annotation to make it clear that the method overrides a method in the superclass, or implements a method in an interface. If your method isn't actually an override, then you'll get a compilation error ... which would be a good thing.)
EDIT
Based on the code that you have added to the question, the problem is not overload vs override. (As I said, I was only guessing ...)
It is most likely that the compareTo and equals are incorrect. It is still not entirely clear exactly where the bug is because the semantics of both methods depends on the compareTo and equals methods of the Boeking class.
The first if statement of the Ticket.compareTo looks highly suspicious. It looks like the return 1; could cause t1.compareTo(t2) and t2.compareTo(t1) to both return 1 for some tickets t1 and t2 ... and that would definitely be wrong.