How to compare array in generic methods in java?

Question

I've got a class using two generic parameters in its constructor. I would like to implement a method to compare two instances of this class, as done in the code below.

It works quite well except when one of the parameter is an arrays. I've read different topics about how to compare arrays with java.util.Arrays.equals(Object[] a, Object[] a2) but as my parameters are not defined as arrays, I can't use this method.

How could return "true" in the content of my array is the same? Or is it possible to cast my parameters to use java.util.Arrays.equals(Object[] a, Object[] a2)?

public class Pair<U, V> {

    public final U first;
    public final V second;

    public Pair(U first, V second) {
        this.first = first;
        this.second = second;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o)
            return true;
        if (o == null || getClass() != o.getClass())
            return false;

        Pair<?, ?> myPair= (Pair<?, ?>) o;

        if (!first.equals(myPair.first) || !second.equals(myPair.second))
            return false;
        return true;
    }
}

Could you define Point as an interface and have different implementations for it? One implementation could be PointOfArrays which would take two arrays. — LuCio
– LuCio, Commented Nov 2, 2018 at 10:26

Oleg Cherednik · Accepted Answer · 2018-11-02 17:58:01Z

1

@Override
public boolean equals(Object o) {
    if (this == o)
        return true;
    if (o == null || getClass() != o.getClass())
        return false;

    Pair<?, ?> myPair = (Pair<?, ?>)o;

    if (first.getClass().isArray() ^ first.getClass().isArray() || second.getClass().isArray() ^ second.getClass().isArray())
        return false;
    return deepEquals(first, myPair.first) && deepEquals(second, myPair.second);
}

private static boolean deepEquals(Object one, Object two) {
    if (one instanceof byte[])
        return Arrays.equals((byte[])one, (byte[])two);
    if (one instanceof char[])
        return Arrays.equals((char[])one, (char[])two);
    if (one instanceof short[])
        return Arrays.equals((short[])one, (short[])two);
    if (one instanceof int[])
        return Arrays.equals((int[])one, (int[])two);
    if (one instanceof long[])
        return Arrays.equals((long[])one, (long[])two);
    if (one instanceof boolean[])
        return Arrays.equals((boolean[])one, (boolean[])two);
    if (one instanceof float[])
        return Arrays.equals((float[])one, (float[])two);
    if (one instanceof double[])
        return Arrays.equals((double[])one, (double[])two);
    if (one instanceof Object[])
        return Arrays.equals((Object[])one, (Object[])two);
    return one.equals(two);
}

edited Nov 2, 2018 at 17:58

answered Nov 2, 2018 at 11:15

Oleg Cherednik

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4 Comments

MC Emperor Over a year ago

It's better to provide some explaination of how this fixes OP's problem. A code-only answer hardly helps future readers.

Pierre GUILLAUME Over a year ago

did you try this code? I've got an Exception when I try to parse (Object[])first

Oleg Cherednik Over a year ago

Yes. What input data do you use?

Pierre GUILLAUME Over a year ago

I use this int[] test = {1, 3, 5}; (and test2 for the second Pair)

GhostCat · Accepted Answer · 2018-11-02 12:37:42Z

0

You can do this:

use the guidance given here to determine whether both objects (this and the "other") contain arrays. Of course, detecting the actual member type means even more work.
utilize Array.equals() to write specific comparison code when the fields are in fact arrays.

But make no mistake: getting your equals() implementation to be correct, and work with inheritance and what not ... will be a major challenge!

For example: do you consider an array of int to be equal to a list of Integer values (assuming the values are all in fact the same)? Or what about nested arrays?

The real point here: Java doesn't have deep equality for arrays, and it is close to impossible to cover all potential cases yourself!

edited Nov 2, 2018 at 12:37

answered Nov 2, 2018 at 10:26

GhostCat

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4 Comments

Michael Over a year ago

That's not "doing two things" - as in two approaches. That's one approach that has 2 steps.

Pierre GUILLAUME Over a year ago

I actually tried to do this, but as first and second are not declared as Array, I've got this error when I try to use Arrays.equals() : The method equals(Object) in the type Object is not applicable for the arguments (V, capture#12-of ?)

Pierre GUILLAUME Over a year ago

That's why I asked if a cast could be a solution but I don't know the type of my array

GhostCat Over a year ago

Casting would not help with anything. See my updated answer.

Adarsh D · Accepted Answer · 2020-03-04 18:59:14Z

public class Fourth {

    public static void main(String[] args) {

        Integer a[] = {1,2,3,4,5};
        Integer b[] = {1,2,3,4,6,7};

        CompareArrays<Integer, Integer> comp = new CompareArrays<Integer, Integer>(a,b);

        comp.compareArrays();

    }
}

class CompareArrays<One extends Number, Two extends Number> {

    One[] array1;
    Two[] array2;

    CompareArrays(One[] ob1, Two[] ob2){
        array1 = ob1;
        array2 = ob2;
    }

    void compareArrays() {

        if(array1.length != array2.length) {
            System.out.println("Two arrays are not equal");

        }else {

            int flag = 0;

            for (int i = 0; i < array1.length; i++) {


                  if(array1[i] != array2[i]) {
                      flag =1;
                  }
            }

            if(flag == 0) {
                System.out.println("Two arrays are equal");
            }else {
                System.out.println("Two arrays are not equal");
            }

        }
    }
}

Michael · Accepted Answer · 2018-11-02 10:40:33Z

It's worth noting that no collection in Java that I'm aware of - pairs, lists, sets... - makes a concession for this case. As an example:

List<Integer[]> one = Arrays.asList(
    new Integer[]{1}, new Integer[]{2}
);
List<Integer[]> two = Arrays.asList(
    new Integer[]{1}, new Integer[]{2}
);
System.out.println(one.equals(two));

Output: false

Equality of two collections should only be determined by the equality of its elements. Anything other than that just makes the contract of your class increasingly vague.

Arrays in Java do not override equals. That's a design decision that's already been made. It's not your job to try to correct what you must perceive as the wrong decision by hacking your implementation.

So, in short, you don't.

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