软件测试第二次作业

　　题目：

　　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.

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
                    

1. Identify the fault. 

　　在遍历数组x的时候，循环条件应该是“i >= 0”,否则遍历不到数组x的第一个数字

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

　　test:x = [1,3,5]; y = 3 

　　Expected = 1

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

　　test:x = [1,3,5]; y = 7 

　　Expected = -1

4.If possible identify a test case that results in an error, but not a failure.

　　test:x = [1,3,5]; y = 1 

　　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

1. Identify the fault. 

　　因为是寻找最后一个0，所以要从后往前找，遍历条件改为“int i = x.length-1; i >=0 ; i--”

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

　　test:x = [0,1,3]

　　Expected = 0

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

　　test:x = [1,3,5]

　　Expected = -1

4.If possible identify a test case that results in an error, but not a failure.

　　test:x = [0,1,0] 

　　Expected = 2