Convert bitwise operation (i & 1) from C++ to Java

Question

I have the following code in C++ from a book/tutorial on OpenGL:

    for (int i = 0; i < 256; i++)
    {
        droplet_x_offset[i] = random_float() * 2.0f - 1.0f;
        droplet_rot_speed[i] = (random_float() + 0.5f) * ((i & 1) ? -3.0f : 3.0f);
        droplet_fall_speed[i] = random_float() + 0.2f;
    }

Now I have tried to convert it to Java, and mostly it is trivial:

    Random random = new Random();
    for (int i = 0; i < 256; i++) {
        dropletXSpeed[i] = random.nextFloat() * 2.0f - 1.0f;
        dropletRotSpeed[i] = (random.nextFloat() + 0.5f) * ((i & 1) ? -3.0f : 3.0f);
        dropletFallSpeed[i] = random.nextFloat() + 0.2f;
    }

Except that ((i & 1) ? -3.0f : 3.0) is not accepted by Java with the message:
incompatible types: int cannot be converted to boolean.

Can someone explain me what (i & 1) exactly does and how should it be replaced in Java?

Answer 1

Unlike C/C++, in Java conditionals have to evaluate to a boolean.

For example:

int i = 1;
if (i) { ... }

isn't valid in java.

Your ternary would need to be:

(i & 1) != 0 ? -3.0f : 3.0f;

Answer 2

Using a trigraph is not as efficient as a formula for simple cases. This is because a branch miss is so expensive. You can do the same like this.

(3 - 6 * (i & 1))

This will be 3 if i is even and -3 if it is odd.

Note: as you have a loop there is an even simpler solution.

for (int i = 0; i < 256; i += 2) {
    dropletXSpeed[i] = random.nextFloat() * 2.0f - 1.0f;
    dropletXSpeed[i+1] = random.nextFloat() * 2.0f - 1.0f;
    dropletRotSpeed[i] = (random.nextFloat() + 0.5f) * 3.0f;
    dropletRotSpeed[i+1] = (random.nextFloat() + 0.5f) * -3.0f;
    dropletFallSpeed[i] = random.nextFloat() + 0.2f;
    dropletFallSpeed[i+1] = random.nextFloat() + 0.2f;
}

BTW nextFloat() is reasonably expensive. If you don't need 24-bit of randomness you can try something like this.

public float next8bitFloat() {
    return random.nextInt(256) * (1.0 / 256);
}