Can (a==1 && a==2 && a==3) evaluate to true in Java?

Question

We know it can in JavaScript.

But is it possible to print "Success" message on the condition given below in Java?

if (a==1 && a==2 && a==3) {
    System.out.println("Success");
}

Someone suggested:

int _a = 1;
int a  = 2;
int a_ = 3;
if (_a == 1 && a == 2 && a_ == 3) {
    System.out.println("Success");
}

But by doing this we are changing the actual variable. Is there any other way?

Answer 1

Yes, it's quite easy to achieve this with multiple threads, if you declare variable a as volatile.

One thread constantly changes variable a from 1 to 3, and another thread constantly tests that a == 1 && a == 2 && a == 3. It happens often enough to have a continuous stream of "Success" printed on the console.

(Note if you add an else {System.out.println("Failure");} clause, you'll see that the test fails far more often than it succeeds.)

In practice, it also works without declaring a as volatile, but only 21 times on my MacBook. Without volatile, the compiler or HotSpot is allowed to cache a or replace the if statement with if (false). Most likely, HotSpot kicks in after a while and compiles it to assembly instructions that do cache the value of a. With volatile, it keeps printing "Success" forever.

public class VolatileRace {
    private volatile int a;

    public void start() {
        new Thread(this::test).start();
        new Thread(this::change).start();
    }

    public void test() {
        while (true) {
            if (a == 1 && a == 2 && a == 3) {
                System.out.println("Success");
            }
        }
    }

    public void change() {
        while (true) {
            for (int i = 1; i < 4; i++) {
                a = i;
            }
        }
    }

    public static void main(String[] args) {
        new VolatileRace().start();
    }
}

Answer 2

Using concepts (and code) from a brilliant code golf answer, Integer values can be messed with.

In this case, it can make ints casted to Integers be equal when they wouldn't normally be:

import java.lang.reflect.Field;

public class Test
{
    public static void main(String[] args) throws Exception
    {
        Class cache = Integer.class.getDeclaredClasses()[0];
        Field c = cache.getDeclaredField("cache");
        c.setAccessible(true);
        Integer[] array = (Integer[]) c.get(cache);
        // array[129] is 1
        array[130] = array[129]; // Set 2 to be 1
        array[131] = array[129]; // Set 3 to be 1

        Integer a = 1;
        if(a == (Integer)1 && a == (Integer)2 && a == (Integer)3)
            System.out.println("Success");
    }
}

Unfortunately it's not quite as elegant as Erwin Bolwidt's multithreaded answer (as this one requires Integer casting), but still some fun shenanigans take place.

Answer 3

In this question @aioobe suggests (and advise against) the use of C preprocessor for Java classes.

Although it is extremely cheaty, that's my solution:

#define a evil++

public class Main {
    public static void main(String[] args) {
        int evil = 1;
        if (a==1 && a==2 && a==3)
            System.out.println("Success");
    }
}

If executed using the following commands it will output exactly one Success:

cpp -P src/Main.java Main.java && javac Main.java && java Main