primitives vs wrapper class initialization

Question

What is the Difference between declaring int's as Below. What are the cases which suits the usage of different Types

int i     = 20;
Integer i = 20;
Integer i = new Integer(20);

Please Note : I have goggled and found that first is going to create primitive int.Second is going to carry out auto boxing and Third is going to create reference in memory.

I am looking for a Scenario which clearly explains when should I use first, second and third kind of integer initialization.Does interchanging the usage is going to have any performance hits

Thanks for Reply.

Answer 1

The initialization in the 1st case is a simple assignment of a constant value. Nothing interesting... except that this is a primitive value that is being assigned, and primitive values don't have "identity"; i.e. all "copies" of the int value 20 are the same.

The 2nd and 3rd cases are a bit more interesting. The 2nd form is using "boxing", and is actually equivalent to this:

Integer i = Integer.valueOf(20);

The valueOf method may create a new object, or it may return a reference to an object that existed previously. (In fact, the JLS guarantees that valueOf will cache the Integer values for numbers in the range -128..+127 ...)

By contrast new Integer(20) always creates a new object.

This issue with new object (or not) is important if you are in the habit of comparing Integer wrapper objects (or similar) using ==. In one case == may be true if you compare two instances of "20". In the other case, it is guaranteed to be false.

The lesson: use .equals(...) to compare wrapper types not ==.

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On the question of which to use:

• If i is int, use the first form.
• If i is Integer, the second form is best ... unless you need an object that is != to other instances. Boxing (or explicitly calling valueOf) reduces the amount of object allocation for small values, and is a worthwhile optimization.

Answer 2

Primitives will take default values when declared without assignment.

But wrapper classes are reference types, so without assignment they will be null. This may cause a NullPointerException to be thrown if used without assignment.