Java autoboxing rules

lkahtz picture lkahtz · Sep 24, 2012 · Viewed 8.6k times · Source

I am a java novice and so confused by the following example. Is it okay to think that "==" sign will compare the values between Integers and "autoboxed" Integers from int, and compare reference address between Integers?

What about doubles and 0/0?

import edu.princeton.cs.introcs.*;

public class Autoboxing {

    public static void cmp(Integer first, Integer second) {
        if (first < second)
            StdOut.printf("%d < %d\n", first, second);
        else if (first == second)
            StdOut.printf("%d == %d\n", first, second);
        else if (first > second)
            StdOut.printf("%d > %d\n", first, second);
        else
            StdOut.printf("%d and %d are incomparable\n", first, second);
    }

    public static void main(String[] args) {
        cmp(new Integer(42), 43);
        cmp(new Integer(42), new Integer(42));
        cmp(43, 43);
        cmp(142, 142);

        Integer a0 = 1000;
        int b0 = 1000;
        Integer c0 = 1000;
        StdOut.println("a0==b0?" + (a0==b0));
        StdOut.println("a0==c0?" + (a0==c0));
        StdOut.println("b0==c0?" + (b0==c0));

        double x1 = 0.0, y1 = -0.0;
        Double a1 = x1, b1 = y1;
        StdOut.println(x1 == y1);
        StdOut.println(a1.equals(b1));

        double x2 = 0.0/0.0, y2 = 0.0/0.0;
        Double a2 = x2, b2 = y2;
        StdOut.println(x2 != y2);
        StdOut.println(!a2.equals(b2));
    }

}

The result is:

42 < 43
42 and 42 are incomparable
43 == 43
142 and 142 are incomparable
=====
a0==b0?true
a0==c0?false
b0==c0?true
=====
true
false
=====
true
false

Answer

lichengwu picture lichengwu · Sep 24, 2012

Unboxing will be happing when arithmetic operators, comparison operators appear.

eg:

Integer a = 10;
a = a+10; //1.unboxing a to int 2.calculate a+10 3.boxing 20 to Integer.
System.out.print(a > 10); //1.unboxing a to int 2. compare

But when == appear, it depends.

If boxing type appear on both side, it will compare the reference.But if base type appear on one side, and the other side is a boxing type, the boxing type will unboxing to base type.

eg:

Integer a = new Integer(129);
Integer b = new Integer(129);
System.out.println(a == b); // compare reference return false
System.out.println(a == 129); // a will unboxing and compare 129 == 129 return true

PS: In Java.lang.Integer Cache to support the object identity semantics of autoboxing for values between -128 and 127 (inclusive) as required by JLS. See source code

So:

Integer a = 127;
Integer b = 127; //cached, the same as b a==b return ture

Integer c = 129;
Integer d = 129; // not cached, c==d return false