Array.equal() giving wrong output

Question

To my understanding, the following code should print true, since both elements are equal.

From java docs Array.get() will return:

Returns the value of the indexed component in the specified array object. The value is automatically wrapped in an object if it has a primitive type.

However, when I run the following code it is printing false:

  public class Test1 {

    static boolean equalTest(Object array1, Object array2) {
        return Array.get(array1, 0).equals(Array.get(array2, 0));
    }

    public static void main(String[] args) {
        int[] a = new int[1];
        byte[] b = new byte[1];
        a[0] = 3;
        b[0] = 3;
        System.out.println(equalTest(a, b));
    }
}

My question is isn't classes implementing Number are or should be directly comparable to one another.

imho it should print false, since they're not the same type. — Jorn Vernee, Jun 23 '17 at 14:00
`Array.get` returns an `Object`: `Integer.valueOf(3).equals(Byte.valueOf((byte) 3))` is false too. — Andy Turner, Jun 23 '17 at 14:01
@Animal it's [`Object`](https://docs.oracle.com/javase/7/docs/api/java/lang/reflect/Array.html#get(java.lang.Object,%20int)). — Andy Turner, Jun 23 '17 at 14:02
@JoshuaTree int is 32 bit and byte is 8 bit...that's what's causing the error — Akshay, Jun 23 '17 at 14:03

score 11 · Accepted Answer · answered Jun 23 '17 at 14:01

11

This has nothing to do with arrays really. Your comparison is equivalent to:

Object x = Integer.valueOf(3);
Object y = Byte.valueOf((byte) 3);
boolean equal = x.equals(y);

That's never going to return true.

Even though your original arrays are of the primitive types, Array.get returns Object, so you're getting the boxed types - and comparing values of those different types.

answered Jun 23 '17 at 14:01

Jon Skeet

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I"m curious, does y have a legible value? – JoshuaTree Jun 23 '17 at 14:03
1

@JoshuaTree: What do you mean by "legible"? It's entirely valid - it's the byte 3 (in `Byte` form). – Jon Skeet Jun 23 '17 at 14:03
@JonSkeet---SIR thanks a lot .... :)Your answers right bang on target...:) – T-Bag Jun 23 '17 at 14:04
@JonSkeet I did'nt quite get why it dosen't return true...as in if it returns object? – Akshay Jun 23 '17 at 14:07
@AndyTurner As in not simpler or the equality check was reversed? If it's the latter, equality should be commutative so it doesn't matter – Michael Jun 23 '17 at 14:08
@Michael reversed. `equals` should be *symmetrical*, but there's no guarantee that it is in general (of course, it is in this case). – Andy Turner Jun 23 '17 at 14:12
@Akshay: It's comparing objects of two different execution-time types (`Byte` and `Integer`). That *should* return `false`. – Jon Skeet Jun 23 '17 at 14:13

score 2 · Answer 2 · answered Jun 23 '17 at 14:09

According to the documentation of Array.get(Object array,int index) method, the value returned is automatically wrapped in an object if it has a primitive type. So, if you add the following lines:

System.out.println(Array.get(array1, 0).getClass());
System.out.println(Array.get(array2, 0).getClass());

you will see the output is

class java.lang.Integer
class java.lang.Byte

The equals method of Integer class first of all checks if the object it is being compared to is also an instance of Integer, if not , then no further checks are required, they are not equal. That's the reason you see the output is false as the objects being compared for equality are Integer and Byte.

score 1 · Answer 3 · edited Jun 23 '17 at 16:06

1

The Array.get invocations return Integer and Byte object instances. These are not equal according to Integer.equals because the class type differs.

edited Jun 23 '17 at 16:06

discipliuned

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answered Jun 23 '17 at 14:02

C-Otto

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sorayadragon · Answer 4 · 2017-06-23T14:06:58.837

0

The array data types don't match. One is int and the other is byte. Since they're being passed to the function as Objects, they'll end up being Integer and Byte. So in order to compare them properly, you need to type-cast them to the same type. Something like:

static boolean equalTest(Object array1, Object array2) {
  int int1 = (int) Array.get(array1, 0);
  int int2 = (int) Array.get(array2, 0);
  return int1.equals(int2);
  // OR return int1 == int2;
}

edited Jun 23 '17 at 14:06

answered Jun 23 '17 at 14:01

sorayadragon

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Well, they're not `int` and `byte`, because `Array.get` returns `Object` - they'll be the wrapper types. (But yes, it's because the types aren't the same.) – Jon Skeet Jun 23 '17 at 14:01
Jon Skeet and I agree. Then we must both be right. :D @jonskeet – Vucko Jun 23 '17 at 14:02
True, but auto-boxing should take care of it. – sorayadragon Jun 23 '17 at 14:04
That's my point though - the types *being compared* are `Integer` and `Byte`, not `int` and `byte`. Casting directly to `int` will give a `ClassCastException`, too - you'd need to cast to `byte` and then convert that `byte` to `int` (whether implicitly or explicitly). – Jon Skeet Jun 23 '17 at 14:06

score 0 · Answer 5 · answered Jun 23 '17 at 14:01

0

The problem is that the Array.get() method returns an object, so for int it'll return Integer, and for byte, it will return Byte. So .equals() method will return false because it first sees whether the type is the same, and then compares the values.

answered Jun 23 '17 at 14:01

Vucko

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score 0 · Answer 6 · answered Jun 23 '17 at 14:17

If you look at the JavaDoc for Array.get you'll see:

The value is automatically wrapped in an object if it has a primitive type.

So your bytes become Bytes and your ints become Integers.

That means the function you're calling is Integer.equals(Object)

This is implemented like so:

public boolean equals(Object obj) {
    if (obj instanceof Integer) {
        return value == ((Integer)obj).intValue();
    }
    return false;
}

You're not passing it an Integer, you're passing it a Byte so that's why it returns false.

Array.equal() giving wrong output

6 Answers6