homework2

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

 

Questions：
1. Identify the fault.

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

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

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

 

回答：

1.

第一个函数没有考虑到x[0]，第二个函数找到第一个为零的值就返回

2.

函数1:

x =null,y=0 ,　　期望值=-1

函数2：

x = null ,期望值=-1

3.

函数1：x=[1,2,3] ,y = 3,期望值=3

函数2：x = [0,1,2]，期望值=0

4.

函数1：x=[1,2,3] ,y = 2,期望值=1

函数2：x = [0,1,2]，期望值=0