Software testing foundations homework2

2018-06-17 21:04:03来源:未知 阅读 ()

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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.

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

56

Questions

? Identify the fault.
? If possible, identify a test case that does not execute the

fault. (Reachability)

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

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

? Due Date: 23:59:59 March 15.

? Please send your answer to tjuscsst@qq.com and post it to your blog.

解:
 
1.
1)for循环里的条件应为i>=0,否则到0直接return-1了.
2)test: x=[]; y = 2  如果是空数组,那就不进入for循环也就不执行,fault了
3)test: x=[2, 3, 5]; y = 5  进入for循环了但是没有循环到最后一次,结果也没有影响,所以执行fault了但是没有error
4)test: x=[2, 3, 5]; y = 9  循环到了最后,少一次,有error,但是因为没有要找的数,所以也不会有failure
2.
1)for循环应该从数组最后一位开,否则找到的是第一个0
2)test: x=[]  空数组,不执行fault
3)test: x=[1]  进入for循环,执行了fault,但是只有一个元素,循环没有正反向之分,所以没有error
4)test: x=[0, 1, 1]  循环反了,有error,但是0的位置都是一样的,所以没有failure

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