TJUSCSST第二次作业

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作业题目

Below are four 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

问题

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.

作业解答：

1.第一个程序的For循环应该i大于等于0，

第二个程序应改为for (int i=x.length-1; i >= 0; i--) ；

2.第一个程序 x = [],  y = 2，

第二个程序 无论怎么输入 都会execute the fault；

3.第一个程序  x = [2,3,4]，y = 4，

第二个程序 test: x = [3]；

4 第一个程序 test: x = [2]，

第二个程序 test: x = [1,0,3]。

TJUSCSST第二次作业

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原文地址：http://www.cnblogs.com/moody/p/5257930.html