Finding the correct partitions for testing digits

Question

I have a function that returns: if x=2y is true., x and y are both digits. A sample code is below:

public class MyClass {

  static boolean xDoubleY(int x, int y) {
    boolean result;
    int temp;

    temp = y * 2;
    if (x == temp) {
        result = true;
        System.out.println("X is twice Y");
    } else {
        result = false;
        System.out.println("X is not twice Y");
    }
    return result;
  }

  public static void main(String[] args) {       
    boolean result1,result2;
    int x1=4;
    int y1=2;
    int x2=5;

    result1 = xDoubleY(x1, y1);
    result2 = xDoubleY(x2, y1);      
 }
}

On one hand, I can think of values for each x and y which are less than, equal or more than zero.

On the other hand, I can think of relation of x and y, and which values would make the expression invalid/valid.

What would be the correct way of finding the correct partitions for this case?

Also, how can boundary value analysis be performed since we have two variables?

Answer 1

You will have three possible outputs:

• True:
  • this will depend on both X and Y. You can pick boundary values for X => { -1, 0, 1 } and its associated values for Y. You do the same for Y => {-1, 0, 1}.
• False:
  • The same as above, but the "associated" value should yield a False result.
• Invalid:
  • It seems to me that the only way to raise an exception is to create an overflow on y*2. Then, you can use y with { (Max int / 2) - 1, (Max int / 2), ( Max int / 2 ) + 1 }.

The values {(Max int / 2) - 1, (Max int / 2)} for y are actually valid - so you can create pairs with x values that will return true and false.