Wrapper equality

[Aniket Patil]

Accroding to the rules of Wrappers,

Integer i1 = 127;
Integer i2 = 127;

i1 == i2, returns True.

However,
Integer i1 = new Integer(127);
Integer i2 = 127;

OR

Integer i1 = new Integer(127);
Integer i2 = new Integer(127);

i1 == i2, returns False in both cases.

Is this because using new( ) creates a fresh object on the heap, ignoring another object with the same value that might exist on the heap?
[ September 11, 2006: Message edited by: Aniket Patil ]

[wise owen]

Why does the autoboxing conversion sometimes return the same reference?

[Barry Gaunt]

The API tells you:

Integer

public Integer(int value)

Constructs a newly allocated Integer object that represents the specified int value.

Parameters:
value - the value to be represented by the Integer object.

You might like to comare it with valueOf:

valueOf

public static Integer valueOf(int i)

Returns a Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor Integer(int), as this method is likely to yield significantly better space and time performance by caching frequently requested values.

Parameters:
i - an int value.
Returns:
a Integer instance representing i.
Since:
1.5

[Burkhard Hassel]

Hi all,


valueOf can be tricky as well:

Integer i1 = Integer.valueOf(42);
Integer i2 = Integer.valueOf("42");
System.out.println(i1==i2);

prints false.
If both parameters were integer literals, it would be true.

Yours,
Bu.

[Meena R. Krishnan]

 
Integer i1 = Integer.valueOf("42");
Integer i2 = Integer.valueOf("42");
System.out.println(i1==i2);      //false   

print false. How?

[Barry Gaunt]

Originally posted by M.R.Krishnan:

 
Integer i1 = Integer.valueOf("42");
Integer i2 = Integer.valueOf("42");
System.out.println(i1==i2);      //false   

print false. How?

Have you checked the API? The caching version of valueOf was recently added in Java 5.0.