软件测试homework2

Identify the fault.

函数findLast,错误得到结果是-1，因为i>0，在i=1时就结束，所以修改为for(int i = x.length-1; i >=0 ; i--)

函数lastZero，错误，得到结果0；因为它找到第一个0就结束，所以修改为for(int i = x.length - 1; i >= 0; i--)

 

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

函数findLast  test : x = [ ] ; y = 2;

函数lastZero test : x = [ ];

 

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

函数findLast  test : x = [1, 2, 3] ; y = 2; except =1;

函数lastZero test : x = [0, 1, 2] ;except = 0;

 

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

函数findLast  test : x = [1, 2, 3] ; y = 4;  返回 -1；

函数lastZero test : x = [1, 2 , 3] ;返回 - 1；