Question

Why does the code below return false for long3 == long2 comparison even though it’s literal.

public class Strings {

    public static void main(String[] args) {
        Long long1 = 256L + 256L;
        Long long2 = 512L;
        Long long3 = 512L;
        System.out.println(long3 == long2);
        System.out.println(long1.equals(long2));
    }
}

Answer 1

Long is an object, not a primitive. By using == you’re comparing the reference values.

You need to do:

if(str.equals(str2))

As you do in your second comparison.

Edit: I get it … you are thinking that other objects act like String literals. They don’t*. And even then, you never want to use == with String literals either.

(*Autobox types do implement the flyweight pattern, but only for values -128 -> 127. If you made your Long equal to 50 you would indeed have two references to the same flyweight object. And again, never use == to compare them. )

Edit to add: This is specifically stated in the Java Language Specification, Section 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.

Note that long is not specifically mentioned but the current Oracle and OpenJDK implementations do so (1.6 and 1.7), which is yet another reason to never use ==

Long l = 5L;
Long l2 = 5L;
System.out.println(l == l2);
l = 5000L;
l2 = 5000L;
System.out.println(l == l2);

Outputs:

true
false