3

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.

Java Beginner
  • 1,635
  • 11
  • 29
  • 51
  • 2
    related: [When to use wrapper class and primitive type](http://stackoverflow.com/questions/1570416/when-to-use-wrapper-class-and-primitive-type), [What is boxing and unboxing and what are the trade offs?](http://stackoverflow.com/questions/13055/what-is-boxing-and-unboxing-and-what-are-the-trade-offs), [New Integer vs valueOf](http://stackoverflow.com/questions/2974561/new-integer-vs-valueof). – Paul Bellora Apr 18 '13 at 04:31

3 Answers3

6

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 ==.

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.
Stephen C
  • 698,415
  • 94
  • 811
  • 1,216
  • Well said Thanks for Reply – Java Beginner Apr 18 '13 at 04:43
  • Integer I is it equal to int i in terms of creation? Will Integer I is not created in the heap memory? As I have got Short s=20 and when I compare Integer I==Short s it will give compile time error but when its int i== Short s its true!!!!! @Stephen C – Gpar Nov 09 '14 at 09:11
  • @Gpar - 1) Unclear what you are asking 2) Unclear what you are asking. 3) That is not a question. – Stephen C Nov 09 '14 at 10:49
  • @Gpar - Hint: if you want proper answers to your questions then: 1) search StackOverflow to see if they have already been asked and answered, then 2) take the time to write a proper Question. – Stephen C Nov 09 '14 at 10:51
3

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.

Paul Bellora
  • 54,340
  • 18
  • 130
  • 181
NPKR
  • 5,368
  • 4
  • 31
  • 48
1

One such scenario I can think of is when you are mapping DB types in Hibernate. If you use Integer you can check for null (assuming the column allows null values). If you use primitive and if the value is null in the database, I guess it throws an error.

VPK
  • 3,010
  • 1
  • 28
  • 35
Sudhanshu Umalkar
  • 4,174
  • 1
  • 23
  • 33