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;
    }
}

```

Answer: 
```
@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);
}

```
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.