java: class Integer == operator strange behaviour [duplicate]

Question

public class IntegerVsInt {

    public static void main(String args[])
    {
        int a = 1;
        int b = 1;
        int c = a + b;

        System.out.println(c);
        System.out.println(a == b);

        Integer x = 1;
        Integer y = 1;
        Integer z = x + y;

        System.out.println(z);
        System.out.println(x == y);
    }
}

In the above code I am comparing two int's and two objects of type integer.

When you compare two int's

a == b

I would expect their values to be compared.

However when you compare two Integer's

x == y

I would expect the address of the two object to be compared and then return a false.

I get true in both the cases? Why is this behavior?

Steven Mastandrea · Accepted Answer · 2013-10-06 05:01:16Z

6

The == is testing whether the Integers are the same object. In java, certain small values are required to be cached, as well as others may optionally be cached, which is why the == Object reference evaluates to true.

The snippet from the JLS Spec 5.1.7

If the value p being boxed is true, false, a byte, or a char in the range \u0000 to \u007f, or an int or short number between -128 and 127 (inclusive), then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.

answered Oct 6, 2013 at 5:01

Steven Mastandrea

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Comments

Suresh Atta · Accepted Answer · 2013-10-06 05:04:56Z

3

x == y

is true for values between -128 and 127 due to integer caching.

Try

 Integer x = 130;
 Integer y = 140;

Now compare and see the magic.

From language spec

If the value p being boxed is true, false, a byte, or a char in the range \u0000 to \u007f, or an int or short number between -128 and 127 (inclusive), then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.

Reason:

The behavior will be the desired one, without imposing an undue performance penalty, especially on small devices. Less memory-limited implementations might.

edited Oct 6, 2013 at 5:04

answered Oct 6, 2013 at 4:58

Suresh Atta

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3 Comments

liv2hak Over a year ago

you mean '==' has been overridden in this case to compare the integer values instead of the address of the objects?

Steven Mastandrea Over a year ago

It's not that '==' has been overridden to compare the values, it's that in all cases where you're creating a new Integer(1) Object, there exists only 1 Object and the internals return that Object reference rather than creating a new one.

Suresh Atta Over a year ago

@liv2hak No not at all the concept of ==. It's concept of caching and improving language performance.

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