Below are four faulty programs. Each includes a test case that results in
failure. Answer the following questions (in the next slide) about each program.
public int findLast (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;
}
// test: x=[2, 3, 5]; y = 2
// Expected = 0
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
Q1 Identify the fault.
Q2 If possible, identify a test case that does not execute the fault. (Reachability)
Q3 If possible, identify a test case that executes the fault, but does not result in an error state.
Q4 If possible identify a test case that results in an error, but not a failure.
A1：findLast函数fault在于循环条件中应为i>=0 lastZero 中fault在于返回值应初设为-1，遍历中改变返回值，循环结束返回返回值
A2：findLast // test: x=[2]; y = 2
// Expected = 0
lastZero // test: x=[ ]
// Expected = -1
A3：findLast // test: x=[2,3,5]; y = 3
// Expected = 1
lastZero // test: x=[ 0,1,2]
// Expected = 0
A4: findLast // test: x=[2,3,5]; y = 3
// Expected = -1
lastZero // test: x=[ 2,1,0]
// Expected = 2
