《软件测试》第二次练习

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