homework2

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--)//错误
{
if (x[i] == y)
{
return i;
}
}
return -1;
}
// test: x=[2, 3, 5]; y = 2
// Expected = 0

Fault：未搜索到数组第一位 ,数组第1位从0开始。  i > 0  改为 i >= 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++)//错误
{
if (x[i] == 0)
{
return i;
}
} return -1;
}
// test: x=[0, 1, 0]
// Expected = 2

Fault：搜索方向错误，应该从后往前扫描。 

int i = 0; i < x.length; i++ 改为 int i=x.length-1; i >= 0; i--


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

（1）test：x=[],y=1;
（2）test：x=[];
    测试数组都为空，直接抛出异常

2.If possible, identify a test case that executes the fault, but does not result in an error state. 
(1)test:x=[4,5,6],y=5,expected = 1
(2)test:x=[2,3,0]

3.If possible identify a test case that results in an error, but not a failure.
（1）test：x=[1,3,5],y=7 output = -1
（2）test：x=[0,2,3] output = 0