作业一
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）for循环应在i<=0处结束
2）x=[] 则指针为空，无法到达fault
3）x=[1,2,3],y=2 此时fault可到达，没有导致error状态
4）x=[3,4,5],y=2 导致error
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）循环应该改为for(int i = x.length-1 ; i >= 0 ; i++)
2）指针异常，无论怎样都会进入循环，进入循环即进入fault
3）x=[0],此时程序到达fault语句，但是没有导致error状态
4）x=[0,1,2] 导致error