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javaautoboxing
Marwadi
Marwadi

Reputation: 151

Autoboxing - Difference between these two lines?

I would like to know the difference between these two:

Integer I = 30;  // is it autoboxing?

Integer I = new Integer(30);  // I know it is autoboxing

Upvotes: 3

Views: 173

Answers (2)

arshajii
arshajii

Reputation: 129517

There is a difference. The first will be treated as 

Integer I = Integer.valueOf(30);  // autoboxing

which will use an Integer object already stored in an internal cache (this is true when autoboxing values in the range -127 to 128 inclusive). The second will explicitly create a new Integer object.

You can see this for yourself:

Integer I1 = 30;
Integer I2 = 30;

System.out.println(I1 == I2);

true

vs

Integer I1 = new Integer(30);
Integer I2 = new Integer(30);

System.out.println(I1 == I2);

false

In general don't compare Integer objects with == since that tests for reference equality. I was using it here to test if I1 and I2 refer to the same instance. In the first snippet, they do because the Integer corresponding to the int 30 already exists in a cache, and this instance is used and assigned to both I1 and I2. In the second snippet, they do not since we are creating two independent objects via new.

By the way there is no autoboxing involved in the second case, you're just creating a simple new object. There is autoboxing in the first case because an int is being implicitly converted to an Integer.

You can actually see the effects of autoboxing by viewing the compiled bytecode with javap -c MyClass.

Upvotes: 6

A4L
A4L

Reputation: 17595

In the first line you write

Integer I = 30;

At the first look you assign an integer (all number literals without an explicit type postfix are considered to be integers) to an object, you think the compiler should complain about incompatible types. But it does not, then this is where the magic happens -> auto boxing! The compiler automatically boxes the integer 30 into the object I, i.e. it creates an Object which hold the integer value 30 and assigns this to your reference.

In the line 

Integer I = new Integer(30);

You explicitly use the new to create an object. So you don't leave the compiler any chance to do anything. Actually, what made so sure that this is auto boxing??

Besides of that, the jvm caches a range of values for constants to minimize the memory used for those constants an thus increase performance. To make use of this feature you should get instances of such object using the valueOf() method. In that case for same integer value one unique object is returned. If you create it using new then for the same constant you'll get a different object each time.

Upvotes: 2

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