Cast Object to specific Array in Java

Question

I'm having some trouble with this for some reason.

I have two java Objects, both of which can be anything, including primitive and non-primitive arrays.

I need to perform an equals check.

If they are arrays, then I have to check their contents instead of their runtime instance.

So for example, let's say I have a method as such:

/**
     * Returns true if the two parameters are arrays, and they both contain
     * the same content.
     * @param aObject1 An object
     * @param aObject2 An object
     * @return
     */
    private boolean equalsArray( Object aObject1, Object aObject2 ) {
      if(aObject1==null){
        return false;
      }
      if(aObject2==null){
        return false;
      }
      if(!aObject1.getClass().isArray()){
        return false;
      }
      if(!aObject2.getClass().isArray()){
        return false;
      }
      //How do I check if the two arrays here contain the same objects
      //without knowledge of their type???

    }

Note that the arrays can be anything, and will most likely not be Object[], but rather Foo[] or Bar[].

Any suggestions? I can't do Array.equals(Object[],Object[]) because I can't cast to Object[].

Arrays.equals((Object[]) aObject1, (Object[]) aObject2); is the way to go and you can of course cast Object to Object[]. — anubhava
– anubhava, Commented Jul 4, 2011 at 15:37
@anubhava if the parms are primitive arrays the cast to Object[] will fail. — pholser
– pholser, Commented Jul 4, 2011 at 15:46

pholser · Accepted Answer · 2011-07-04 15:41:04Z

2

Use java.lang.reflect.Array:

    if (Array.getLength(first) != Array.getLength(second))
        return false;

    for (int i = 0; i < Array.getLength(first); ++i) {
        Object firstItem = Array.get(first, i);
        Object secondItem = Array.get(second, i);
        if (!(firstItem == null ? secondItem == null : firstItem.equals(secondItem)))
            return false;
    }
    return true;

answered Jul 4, 2011 at 15:41

pholser

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2 Comments

Peter Lawrey Over a year ago

+1: This is likely to be simplest. It will mean an int[] and Integer[] can be equal.

pholser Over a year ago

Or, if you need to handle multi-dimensional arrays correctly, recursively call your method when testing item equality. Or, just use the previously mentioned Commons Lang method.

Lukas Eder · Accepted Answer · 2011-07-04 15:40:37Z

There is a nice equals utility in

boolean org.apache.commons.lang.ArrayUtils.isEquals(Object array1, Object array2)

It will run something like this - which of course you don't want to write manually :-)

public static boolean isEquals(Object array1, Object array2) {
    return new EqualsBuilder().append(array1, array2).isEquals();
}

public EqualsBuilder append(Object lhs, Object rhs) {
    if (isEquals == false) {
        return this;
    }
    if (lhs == rhs) {
        return this;
    }
    if (lhs == null || rhs == null) {
        this.setEquals(false);
        return this;
    }
    Class lhsClass = lhs.getClass();
    if (!lhsClass.isArray()) {
        // The simple case, not an array, just test the element
        isEquals = lhs.equals(rhs);
    } else if (lhs.getClass() != rhs.getClass()) {
        // Here when we compare different dimensions, for example: a boolean[][] to a boolean[] 
        this.setEquals(false);
    }
    // 'Switch' on type of array, to dispatch to the correct handler
    // This handles multi dimensional arrays of the same depth
    else if (lhs instanceof long[]) {
        append((long[]) lhs, (long[]) rhs);
    } else if (lhs instanceof int[]) {
        append((int[]) lhs, (int[]) rhs);
    } else if (lhs instanceof short[]) {
        append((short[]) lhs, (short[]) rhs);
    } else if (lhs instanceof char[]) {
        append((char[]) lhs, (char[]) rhs);
    } else if (lhs instanceof byte[]) {
        append((byte[]) lhs, (byte[]) rhs);
    } else if (lhs instanceof double[]) {
        append((double[]) lhs, (double[]) rhs);
    } else if (lhs instanceof float[]) {
        append((float[]) lhs, (float[]) rhs);
    } else if (lhs instanceof boolean[]) {
        append((boolean[]) lhs, (boolean[]) rhs);
    } else {
        // Not an array of primitives
        append((Object[]) lhs, (Object[]) rhs);
    }
    return this;
}

Oscar Gomez · Accepted Answer · 2011-07-04 15:40:27Z

0

You may use

  Arrays.equals(Object[] a, Object[] a2)

However you will have to implement the equals for the contents since the method does:

 o1.equals(o2)

This will only work if you implement equals for that object.

For non primitive types there is an

 Arrays.equals

override for every primitive type.

answered Jul 4, 2011 at 15:40

Oscar Gomez

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Comments

MarcoS · Accepted Answer · 2011-07-04 15:42:15Z

0

You can use Arrays.equal(..) or, if the arrays can have nested arrays Arrays.deepEquals(..)

answered Jul 4, 2011 at 15:42

MarcoS

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3 Comments

Tovi7 Over a year ago

No it won't. If you have a class Foo[], then this will throw an exception because it can't case Foo[] to Object[]

MarcoS Over a year ago

@Tovi7: I don't understand your comment: a Foo[] is an Object[]: no need for casting

Tovi7 Over a year ago

I have a compile time "Object" which at runtime is "Foo[]". I have to cast to Object[] or the method wont work. If I do cast it to Object[] and execute your methods (like I stated in my original question) I get classcastexceptions in the method.

Eng.Fouad · Accepted Answer · 2011-07-04 15:47:30Z

0

How about:

if(aObject1.getClass().getName().equals(aObject2.getClass().getName()))
return Arrays.equals(aObject1, aObject2);
else return false;

edited Jul 4, 2011 at 15:47

answered Jul 4, 2011 at 15:42

Eng.Fouad

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