题目:
(1)Identify the fault.
(2)If possible, identify a test case that does not execute the fault. (Reachability)
(3)If possible, identify a test case that executes the fault, but does not result in an error state.
(4)If possible identify a test case that results in an error, but not a failure.
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.fault: for循环的终止条件应改为i>=0
2.在给出的例子中,判断到i=0时已经跳出了循环,所以会return -1; 故应该为Excepted: 0 ,Actual: -1;
3.执行faul条件但是没有出现error的样例: x=[1,2,3] , y=2。 得到index=1,执行了判断条件,但是并不会报错 故为Excepted: 1 ,Actual: 1;
4.出现error但无failure的样例:x=[1,2,3] ,y=0。 return -1; 故Excepted: -1 ,Actual: -1;
作业二:
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.应该为 for (int i=x.length-1; i>= 0; i--){}
2.不存在不会执行fault的样例。
3.出现fault但没有error的样例: x=[0] 。此处会找到正确答案,但是程序的逻辑仍然存在错误。Excepted: 0, Actual: 0;
4.出现error但没有failure的样例:x=[1,3,7,0,4,6] 。 由于检索的顺序反了,所以只要是数组内元素个数多于一个,就是处于error。Excepted: 2,Actual: 2