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Possible Duplicate:
Weird Java Boxing

Recently while I was reading about wrapper classes I came through this strange case:

Integer i1 = 1000;
Integer i2 = 1000;

if(i1 != i2) System.out.println("different objects");

if(i1 == i2) System.out.println("same object");

Which prints:

different objects

and 

Integer i1 = 10;
Integer i2 = 10;

if(i1 != i2) System.out.println("different objects");

if(i1 == i2) System.out.println("same object");

Which prints:

same object

Is there any reasonable explanation for this case?

Thanks

Community
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Feras Odeh
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3 Answers3

9

The reason why == returns true for the second case is because the primitive values boxed by the wrappers are sufficiently small to be interned to the same value at runtime. Therefore they're equal.

In the first case, Java's integer cache is not large enough to contain the number 1000, so you end up creating two distinct wrapper objects, comparing which by reference returns false.

The use of said cache can be found in the Integer#valueOf(int) method (where IntegerCache.high defaults to 127):

public static Integer valueOf(int i) {
    if(i >= -128 && i <= IntegerCache.high)
        return IntegerCache.cache[i + 128];
    else
        return new Integer(i);
}

As Amber says, if you use .equals() then both cases will invariably return true because it unboxes them where necessary, then compares their primitive values.

BoltClock
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1

Integer i1 = 1000;

compiler understands it as an int that is why i1 == i2 // return true

but while i1 and i2 are big == test return false

Integer i1 = 10000000;
Integer i2 = 10000000;

if(i1 != i2) System.out.println("different objects"); // true

if(i1 == i2) System.out.println("same object"); // false

notice :

Integer i1 = 100;
Integer i2 = 100;

System.out.println(i1 == i2); // true

i1 = 1000000;
i2 = 1000000;

System.out.println(i1 == i2); // false

do not test equality with == for Objects. it just compares their reference. .equal() check whether two objects are same or not.

Integer i1 = 1000000;
Integer i2 = 1000000;

i1 == i2 // false
i1.equals(i2) // true
ironwood
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  • "compiler understands it as an integer that is why i1 == i2 // return true" This doesn't make any sense as if you tried for example 700 it gives you different objects aren't this small value – Feras Odeh Jan 09 '11 at 07:05
  • I mean int a simple data type not object –  Jan 09 '11 at 07:06
-1

I just tried this, and all it prints for me is

different objects

as expected, since you are creating two different wrapper objects, even though they happen to contain the same value.

As Amber implies in the above comment,

if(i1.equals(i2)) System.out.println("same value");

does indeed print

same value
metadaddy
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