Java autoboxing rules

Question

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

Near

Have you tried? System.out.println("eq " + (new Integer(10) == 10)); – Nica

answer explained here link – Highoctane

Answer 1

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

Betook

But how come does Object o1=3; Object o2=3; System.out.println(o1==o2); return true? Both sides of the == here are boxing types – Darra

This happens because of the Integer Cache. Under the hood, the compiler replaces the assignments above with Object o1=Integer.valueOf(3); and Object o2=Integer.valueOf(3);. The implementation of Integer.valueOf(int) saves certain instances of Integer for re-use. Thus, while the two objects on either side of the == operator are never unboxed, they have the same reference. More: javapapers.com/java/java-integer-cache – Marsha

Answer 2

Here is the Tutorial for Autoboxing and Unboxing.

You can also go through JLS#5.1.7. Boxing Conversion and JLS#5.1.8. Unboxing Conversion

0.0 / 0.0 is NaN you can not compare infinity at least in maths. I guess that is why this comparison does not work.

From JLS #4.2.3. Floating-Point Types, Formats, and Values

Positive zero and negative zero compare equal; thus the result of the expression 0.0==-0.0 is true and the result of 0.0>-0.0 is false

NaN is unordered, so:

• The numerical comparison operators <, <=, >, and >= return false if either or both operands are NaN (§15.20.1).

• The equality operator == returns false if either operand is NaN.

• In particular, (x=y) will be false if x or y is NaN.

• The inequality operator != returns true if either operand is NaN (§15.21.1).

• In particular, x!=x is true if and only if x is NaN.

If you check Double#equals method it has two exceptions

also has the value true. However, there are two exceptions:

• If d1 and d2 both represent Double.NaN, then the equals method returns true, even though Double.NaN==Double.NaN has the value false.

• If d1 represents +0.0 while d2 represents -0.0, or vice versa, the equal test has the value false, even though +0.0==-0.0 has the value true.

This definition allows hash tables to operate properly.

Tolerance

thanks~ so a2 and b2 both are boolean value false, therefore equal to each other? – Near

a2.equals(b2) will always return true if both are NAN which is documented in the Java doc of equals method. – Tolerance

Answer 3

== can only be used for checking if the variables are equal or not if the variables are primitive types. For object variables, == is used to compare the reference of the objects. If you want to compare the values of the objects, use .equals() method.

Phalan

Answer 4

I would not recommend comparing boxed ints with ==, as it works only for some values.

Monosyllable

I am also reading articles, it seems that for int, == compares the values between -127 - 127. – Near

Yes. it's kind of a cache. Use .equals() when you have Integers and == when you have ints. – Monosyllable