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

Answer 4

As we already know that it is possible to make this code evaluate to true thanks to great answers of Erwin Bolwidt and phflack, I wanted to show that you need to keep a high level of attention when dealing with a condition that looks like the one presented in the question, as sometimes what you see might not be exactly what you think it is.

This is my attempt to show that this code prints Success! to the console. I know I cheated a bit, but I still think this is a good place to present it right here.

No matter what the purposes of writing code like this are - better to know how to deal with the following situation and how to check if you're not wrong with what you think you see.

I used the Cyrillic 'a' which is a distinct character from the latin 'a'. You can inspect the characters used in the if statement here.

This works because the names of the variables are taken from different alphabets. They are distinct identifiers, creating two distinct variables with a different value in each.

Note that if you want this code to work properly, character encoding needs to be changed to one supporting both characters, e.g. all Unicode encodings (UTF-8, UTF-16 (in BE or LE), UTF-32, even UTF-7), or Windows-1251, ISO 8859-5, KOI8-R (thank you - Thomas Weller and Paŭlo Ebermann - for pointing it out):

public class A {
    public static void main(String[] args) {
        int а = 0;
        int a = 1;
        if(а == 0 && a == 1) {
            System.out.println("Success!");
        }
    }
}

(I hope you will never have to deal with that sort of problem any time in the future.)

Answer 5

There is another way to approach this (in additional to the volatile data-racing approach that I posted earlier), using the power of PowerMock. PowerMock allows methods to be replaced with other implementations. When that is combined with auto-unboxing, the original expression (a == 1 && a == 2 && a == 3), without modification, can be made true.

@phflack's answer relies on modifying the auto-boxing process in Java that uses the Integer.valueOf(...) call. The below approach relies on modifying auto-unboxing by changed the Integer.intValue() call.

The advantage of the below approach is that the original if-statement given by the OP in the question is used unchanged, which I think is the most elegant.

import static org.powermock.api.support.membermodification.MemberMatcher.method;
import static org.powermock.api.support.membermodification.MemberModifier.replace;

import java.util.concurrent.atomic.AtomicInteger;

import org.junit.Assert;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.powermock.core.classloader.annotations.PrepareForTest;
import org.powermock.modules.junit4.PowerMockRunner;

@PrepareForTest(Integer.class)
@RunWith(PowerMockRunner.class)
public class Ais123 {
    @Before
    public void before() {
        // "value" is just a place to store an incrementing integer
        AtomicInteger value = new AtomicInteger(1);
        replace(method(Integer.class, "intValue"))
            .with((proxy, method, args) -> value.getAndIncrement());
    }

    @Test
    public void test() {
        Integer a = 1;

        if (a == 1 && a == 2 && a == 3) {
            System.out.println("Success");
        } else {
            Assert.fail("(a == 1 && a == 2 && a == 3) != true, a = " + a.intValue());
        }
    }

}

Answer 6

Since this seems to be a follow-up of this JavaScript question, it’s worth noting that this trick and similar works in Java too:

public class Q48383521 {
    public static void main(String[] args) {
        int aﾠ = 1;
        int ﾠ2 = 3;
        int a = 3;
        if(aﾠ==1 && a==ﾠ2 && a==3) {
            System.out.println("success");
        }
    }
}

On Ideone

---

But note that this isn’t the worst thing you could do with Unicode. Using white-space or control characters that are valid identifiers parts or using different letters that look the same still creates identifiers that are different and can be spotted, e.g. when doing a text search.

But this program

public class Q48383521 {
    public static void main(String[] args) {
        int ä = 1;
        int ä = 2;
        if(ä == 1 && ä == 2) {
            System.out.println("success");
        }
    }
}

uses two identifiers that are the same, at least from the Unicode point of view. They just use different ways to encode the same character ä, using U+00E4 and U+0061 U+0308.

On Ideone

So depending on the tool you’re using, they may not only look the same, Unicode enabled text tools may not even report any difference, always finding both when searching. You may even have the problem that the different representations get lost when copying the source code to someone else, perhaps trying to get help for the “weird behavior”, making it non-reproducible for the helper.

Answer 7

Inspired by the @Erwin's excellent answer, I wrote a similar example, but using Java Stream API.

And an interesting thing is that my solution works, but in very rare cases (because  just-in-time compiler optimizes such a code).

The trick is to disable any JIT optimizations using the following VM option:

-Djava.compiler=NONE

In this situation, the number of success cases increases significantly. Here is the code:

class Race {
    private static int a;

    public static void main(String[] args) {
        IntStream.range(0, 100_000).parallel().forEach(i -> {
            a = 1;
            a = 2;
            a = 3;
            testValue();
        });
    }

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

P.S. Parallel streams use ForkJoinPool under the hood, and variable a is shared between multiple threads without any synchronization, that's why the result is non-deterministic.

Answer 8

Along similar lines, by forcing a float (or double) to underflow (or overflow) through division (or multiplication) by a large number:

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