Object type in Java and referencing arrays

Question

public class RefMix {
  public static void main(String[] args) {
    Object[] a = {null, "foo"};
    Object[] b = {"bar", b};
    a[0] = b;

    System.out.println(a[0][0]);
  }
}

My understanding is that arrays are objects in Java, and therefore a subclass of the Object type. My further understanding is that a 2-dim array is implemented as an array of references to arrays. Therefore I don't understand why my a[0][0] does not produce bar in the code above. Instead it doesn't compile:

RefMix.java:7: array required, but java.lang.Object found

Code-Apprentice · Accepted Answer · 2012-10-07 00:03:57Z

13

My understanding is that arrays are objects in Java, and therefore a subclass of the Object type. My further understanding is that a 2-dim array is implemented as an array of references to arrays.

This is all correct and explains why you can do the assignment

a[0] = b;

without any complaints from the compiler.

Therefore I don't understand why my a[0][0] does not produce bar in the code above.

Okay, let's take a look at the types in this expression:

a is a Object[] -- that is an array of Objects
a[0] is an Object
a[0][0] -- Now you are trying to use an array subscript on an Object. The compiler does not know that the Object is in fact an array, so it complains.

answered Oct 7, 2012 at 0:03

Code-Apprentice

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Comments

Antimony · Accepted Answer · 2012-10-07 00:00:09Z

10

The runtime type of an object instance differs from the statically inferred type. The compiler will try to estimate what type each variable could be in the program, in order to catch certain types of errors early. In this case, a[0] will always be an array, but the compiler doesn't know this. Since you are getting an object out of an object array, all the compiler knows is that a[0] is an object. Therefore it raises an error.

In cases where you know something will always be a certain type but the compiler can't figure it out, you can get around this by inserting an explicit cast.

System.out.println(((Object[])a[0])[0]);

answered Oct 7, 2012 at 0:00

Antimony

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4 Comments

Fixee Over a year ago

The ugliness of a cast like this has inspired me to give up using arrays like this; I'll just use a "normal" class of object references. But thanks!

Antimony Over a year ago

@Fixee, arrays in Java are rather ugly, but you can often avoid the need for casts by reducing the number of times you store heterogeneous data. You don't normally need to store both strings and object[]s in the same array.

brimborium Over a year ago

@Antimony I actually can not find a single use case where you would need such a thing... ;)

bestsss Over a year ago

@brimborium, that's bit too much, here is one case: if you have a ORM and cache the data or just need an intermediate layer, it's likely to have object[] w/ all the fields of the target class, then assign properly to the instance. Overall, it's quite common in the middleware. Here is another case invoking anything through reflection takes Object[] as args and the latter can contain whatever necessary.

brimborium · Accepted Answer · 2012-10-07 00:08:14Z

7

You are right, arrays are always Objects in Java, but Objects are not always arrays, therefore you get a compile error, because a is an Object[] (one dimensional). You can not access

a[0][0];

because a is not a two dimensional array (at least it's not declared as such). However in this case, you are sure, that a[0] is an array of Objects. Therefore, you can do this:

Object[] c = (Object[]) a[0];
System.out.println(c[0]);
// or directly:
System.out.println(((Object[])a[0])[0]);

This casts the return type of a[0] (which is Object), into a Object[] and you can then access the "second layer" of the array.

answered Oct 7, 2012 at 0:08

brimborium

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Comments

Rohit Jain · Accepted Answer · 2012-10-07 00:01:06Z

4

Since array in Java is an object, so the 1st and 2nd assignment, will store your array as the 1st element of your Object array..

Object[] a = {null, "foo"};
Object[] b = {"bar", b};

Now, you have changed your 1st element of a object array to contain value b instead of array.. But since it is an array of object. Everything coming out of it will be object..

Since a[0] is an object.. So, you clearly can't access something like this: -

System.out.println(a[0][0]);

You can try to typecast your object a[0] to object array..: -

System.out.println(((Object[])a[0])[0]);

answered Oct 7, 2012 at 0:01

Rohit Jain

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Comments

Sidharth Mudgal · Accepted Answer · 2012-10-07 00:02:06Z

4

Do:

System.out.println(((Object[])a[0])[0]);

This will "cast" the object to an object array at runtime.

answered Oct 7, 2012 at 0:02

Sidharth Mudgal

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Comments

David Lee · Accepted Answer · 2012-10-07 19:34:49Z

As I do a lot of C++ coding lately, I find that Java is a lot more strict in type checking then C++ is. Java sees everything but primitive types as Object but Java goes one step further in distinguishing Array and non-Array types. During assignment like "a[0] = b;" , Java first check to see if it is an Array type or not, after that it goes through regular polymorphic type checking procedure. If you wish to make your code work, you should do...

Object[][] a = {{null}, {"foo"}};
Object[] b = {"bar", new java.util.Date()};
a[0] = b;

You can see how java take special care on array types by looking into Java Class signature that is passed to Class.forName() as parameter. For example, data type ..

com.foo.Bar[][] barsIn2D;

can be loaded with signature below...

// [ => Array
// [[ => Array of Array type
// L<object>; => Object, not Array
Class.forName("[[Lcom/foo/Bar;");

As you see, signature starts with either '[' or 'L'. This tells us whether its an array or not takes precedence over "Lcom/foo/Bar;".

Bhesh Gurung · Accepted Answer · 2012-10-07 00:22:04Z

1

Everything that you are doing is equivalent to this

Object a = {"bar", "foo"};
System.out.println(a[0]);

which doesn't compile either.

answered Oct 7, 2012 at 0:22

Bhesh Gurung

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