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.