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Strange Wrapper Classes behavior with == and != [duplicate]

Tags:

java

wrapper

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

like image 374
Feras Odeh Avatar asked Dec 02 '25 08:12

Feras Odeh


1 Answers

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.

like image 183
BoltClock Avatar answered Dec 03 '25 22:12

BoltClock



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