[ST2017] Hw2: Fault, Error, Failure and test case for projects

3014218071 王漢超

Q: Below are two faulty programs. Each includes a test case that results in failure. Answer the following questions (in the next slide) about each program.

• Identify the fault.
• If possible, identify a test case that does not execute the fault. (Reachability)
• If possible, identify a test case that executes the fault, but does not result in an error state.
• If possible identify a test case that results in an error, but not a failure.

public int findLast (int[] x, int y) {
/* Pro.1
 * Effects: If x==null, throw NullPointerException
 * else return the index of the last element
 * in x that equals y.
 * If no such element exists, return -1
 */
  for (int i=x.length-1; i > 0; i--)
  {
    if (x[i] == y)
    {
      return i;
    }
  }
  return -1;
}
// test: x=[2, 3, 5]; y = 2
// Expected = 0

A1:

The fault: When x != null, x[0] will not be checked, even if only x[0] equals to y;

x = [2, 3, 5]; y = 2; -> return -1 

Test case 1: x = []; y = 2; -> throw NullPointerException, not execute the fault.

Test case 2: x = [0, 1, 2]; y = 2; -> i = 2, return 2, not result in an error state.

Test case 3: x = [0, 1, 3]; y = 2; -> i = 1, return -1, not a failure.

public static int lastZero (int[] x) {
/* Pro.2
 * Effects: if x==null, throw NullPointerException
* else return the index of the LAST 0 in x.
 * Return -1 if 0 does not occur in x
 */
  for (int i = 0; i < x.length; i++)
  {
    if (x[i] == 0)
    {
      return i;
    }
  }
  return -1;
}
// test: x=[0, 1, 0]
// Expect: return 2

A2:

The fault: When x != null, the index of the FIRST 0 in x will be returned but not the LAST 0.

x = [0, 1, 0 ]; -> return 0 

Test case 1: x = []; -> throw NullPointerException, not execute the fault.

Test case 2: x = [1, 2, 3]; -> i = 3, return -1, not result in an error state.

Test case 3: x = [1, 0, 1]; -> i = 1, return 1, not a failure.