2

As far as my understanding goes Object is also a class in java. So,how is it that we can write

Object ob = new Integer[2];

and not 

Integer i = new Integer[2];

How is it that a single reference ob can point to array of Integers but reference of type Integer can't?

Improve this question
6
  • 3
    Object is the mother of all classes -- all other classes inherit from it. Integer and Integer[] are two different classes where neither inherits from the other. Commented Nov 11, 2012 at 15:17
  • ok that explains it, so basically Integer[2] is treated as a different class:) Commented Nov 11, 2012 at 15:18
  • 1
    In Java, array length is not part of the type. There is no type called Integer[2]. The type of arrays of Integer objects is written Integer[]. Commented Nov 11, 2012 at 15:26
  • 1
    Yep, Integer[] is a distinct class from Integer. The array size is not a part of the class identity, though. (But also understand that Integer[][] is a distinct class from Integer[].) Commented Nov 11, 2012 at 15:53
  • 2
    (It should be noted that this is one place where Java differs significantly from C, since in C "arrays decay to pointers", but in Java that's not the case.) Commented Nov 11, 2012 at 19:17

4 Answers 4

Reset to default
5

Because every array is an object. So polymorphism rules tell us that Integer[] (array of Integer) can be stored in an Object, but obviously Integer[] (array of Integer) can not be stored in an Integer.

Improve this answer
Sign up to request clarification or add additional context in comments.

9 Comments

Your statements are quite confusing. You can store Integer[2] in Integer. You cannot store new Integer[2] in Integer.
@RohitJain what do you mean when you say "You can store Integer[2] in Integer"
I mean, Integer[2] seems like accessing 2nd index in some array names Integer. I know its not possible. But you should edit it to make it more sensible. Just a Suggestion.
@RohitJain: You're confusing types with expressions. The type of of arrays of Integer objects (that is, Integer[].class) is distinct from the type of Integer objects (that is, Integer.class).
@DanielPryden.. That I know. I'm talking about this: - Integer[] Integer = new Integer[2]; System.out.println(Integer[1]); Now Integer[1] is an Integer here. That's why I said, writing new Integer[2] will be more sensible, as it would mean an array rather than Integer[2] that sounds like an array element.. Of course that is hardly goind to be noticed.
|
3

From JLS#Chapter 10. Arrays

In the Java programming language, arrays are objects (§4.3.1), are dynamically created, and may be assigned to variables of type Object (§4.3.2). All methods of class Object may be invoked on an array.

All the components of an array have the same type, called the component type of the array. If the component type of an array is T, then the type of the array itself is written T[].

Improve this answer

1 Comment

thanks this explains a lot:), obviously it was a tough choice to choose the right answer, but the last line in you answer made more sense, thanks all for answering so fast:)
3

Because every array is a subtype of Object. But no array is a subtype of Integer.

Improve this answer

Comments

1

Since the following expression on the RHS creates an array object: -

new Integer[2];

So, your reference type on the LHS should be compatible to be able to hold a reference to an array.

Now, since array in Java is a subtype of Object, so an Object type can hold a reference to an array. But, an Integer reference of course cannot point to an array.

So, 

Object ob = new Integer[2]; // is valid. as `Object` type reference can point to an array object
Integer i = new Integer[2]; // is not valid. as `Integer` type reference cannot point to an array object.
Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.