How to Compare the Java Byte[] Array?
public class ByteArr {

    public static void main(String[] args){
        Byte[] a = {(byte)0x03, (byte)0x00, (byte)0x00, (byte)0x00};
        Byte[] b = {(byte)0x03, (byte)0x00, (byte)0x00, (byte)0x00};
        byte[] aa = {(byte)0x03, (byte)0x00, (byte)0x00, (byte)0x00};
        byte[] bb = {(byte)0x03, (byte)0x00, (byte)0x00, (byte)0x00};

        System.out.println(a);
        System.out.println(b);
        System.out.println(a == b);
        System.out.println(a.equals(b));

        System.out.println(aa);
        System.out.println(bb);
        System.out.println(aa == bb);
        System.out.println(aa.equals(bb));
    }
}

I do not know why all of them print false.

When I run "java ByteArray", the answer is "false false false false".

I think the a[] equals b[] but the JVM is telling me I am wrong, why??

2

Use Arrays.equals() if you want to compare the actual content of arrays that contain primitive types values (like byte).

System.out.println(Arrays.equals(aa, bb));

Use Arrays.deepEquals for comparison of arrays that contain objects.

2

Cause they're not equal, ie: they're different arrays with equal elements inside.

Try using Arrays.equals() or Arrays.deepEquals().

2

As byte[] is mutable it is treated as only being .equals() if its the same object.

If you want to compare the contents you have to use Arrays.equals(a, b)

BTW: Its not the way I would design it. ;)

have you looked at Arrays.equals()?

Edit: if, as per your comment, the issue is using a byte array as a HashMap key then see this question.

2

If you're trying to use the array as a generic HashMap key, that's not going to work. Consider creating a custom wrapper object that holds the array, and whose equals(...) and hashcode(...) method returns the results from the java.util.Arrays methods. For example...

import java.util.Arrays;

public class MyByteArray {
   private byte[] data;

   // ... constructors, getters methods, setter methods, etc...


   @Override
   public int hashCode() {
      return Arrays.hashCode(data);
   }

   @Override
   public boolean equals(Object obj) {
      if (this == obj)
         return true;
      if (obj == null)
         return false;
      if (getClass() != obj.getClass())
         return false;
      MyByteArray other = (MyByteArray) obj;
      if (!Arrays.equals(data, other.data))
         return false;
      return true;
   }


}

Objects of this wrapper class will work fine as a key for your HashMap<MyByteArray, OtherType> and will allow for clean use of equals(...) and hashCode(...) methods.

Try for this:

boolean blnResult = Arrays.equals(byteArray1, byteArray2);

I am also not sure about this, but try this may be it works.

Because neither == nor the equals() method of the array compare the contents; both only evaluate object identity (== always does, and equals() is not overwritten, so the version from Object is being used).

For comparing the contents, use Arrays.equals().

They are returning false because you are testing for object identity rather than value equality. This returns false because your arrays are actually different objects in memory.

If you want to test for value equality should use the handy comparison functions in java.util.Arrays

e.g.

import java.util.Arrays;

'''''

Arrays.equals(a,b);

You can also use a ByteArrayComparator from Apache Directory. In addition to equals it lets you compare if one array is greater than the other.

why a[] doesn't equals b[]? Because equals function really called on Byte[] or byte[] is Object.equals(Object obj). This functions only compares object identify , don't compare the contents of the array.

I looked for an array wrapper which makes it comparable to use with guava TreeRangeMap. The class doesn't accept comparator.

After some research I realized that ByteBuffer from JDK has this feature and it doesn't copy original array which is good. More over you can compare faster with ByteBuffer::asLongBuffer 8 bytes at time (also doesn't copy). By default ByteBuffer::wrap(byte[]) use BigEndian so order relation is the same as comparing individual bytes.

.

Java byte compare,

public static boolean equals(byte[] a, byte[] a2) {
        if (a == a2)
            return true;
        if (a == null || a2 == null)
            return false;

        int length = a.length;
        if (a2.length != length)
            return false;

        for (int i = 0; i < length; i++)
            if (a[i] != a2[i])
                return false;

        return true;
    }

You can also use org.apache.commons.lang.ArrayUtils.isEquals()

Arrays.equals is not enough for a comparator, you can not check the map contain the data. I copy the code from Arrays.equals, modified to build a Comparator.

class ByteArrays{
    public static <T> SortedMap<byte[], T> newByteArrayMap() {
        return new TreeMap<>(new ByteArrayComparator());
    }

    public static SortedSet<byte[]> newByteArraySet() {
        return new TreeSet<>(new ByteArrayComparator());
    }

    static class ByteArrayComparator implements Comparator<byte[]> {
        @Override
        public int compare(byte[] a, byte[] b) {
            if (a == b) {
                return 0;
            }
            if (a == null || b == null) {
                throw new NullPointerException();
            }

            int length = a.length;
            int cmp;
            if ((cmp = Integer.compare(length, b.length)) != 0) {
                return cmp;
            }

            for (int i = 0; i < length; i++) {
                if ((cmp = Byte.compare(a[i], b[i])) != 0) {
                    return cmp;
                }
            }

            return 0;
        }
    }
}

As a newer solution, you can use

import org.junit.jupiter.api.Assertions.*

assertArrayEquals(aa,bb);

There's a faster way to do that:

Arrays.hashCode(arr1) == Arrays.hashCode(arr2)
1
Marcus Vance

Marcus Vance

Cybersecurity & Digital Privacy Researcher

Marcus Vance is a cybersecurity auditor and technology writer dedicated to educating the public about online safety, data privacy regulations, enterprise security, and emerging cyber threats.

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