Homework2

Homework 2

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.

1) If possible, identify a test case that does not execute the fault.(Reachability)

2) If possible, identify a test case that executes the fault, but does not result in an error state.

3) If possible, identify a test case that results in an error， but not a  failure

 

public int findTheLast (int[] x, int y) {

//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;

}

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

 

x = [2, 3, 5]; y = 2;

return -1;

 

Test case 1: x = []; y = 3; -> throw NullPointerException, not execute the fault.

 

Test case 2: x = [0, 1, 3]; y = 3; -> i = 2, return 2, executed the fault, but not result in an error state.

 

Test case 3: x = [0, 1, 3]; y = 2; When i = 1, return -1, it result in an error, but not a failure.

 

 

public static int lastZero (int[] x) {

//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] // Expected = 2

 

The fault: When x != null, there are more than 1 elements in x equaled to 0, the test case will return the first zero, not the last zero;

 

x = [0, 1, 0];

return 0;

 

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

 

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

 

Test case 3: x = [0, 1, 3]; -> i = 0, return 0, it result in an error, but not a failure.