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  • 软件测试homework2

    题目:

    Below are two faulty programs. Each includes a test case that results in failure. Answer the following questions (in the next slide) about each program.

    1.

    public intfindLast(int[] x, inty) {
    //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 (inti=x.length-1; i> 0; i--)
    {
    if (x[i] == y)
    {
    return i;
    }
    }
    return -1;
    }
    // test: x=[2, 3, 5]; y = 2
    // Expected = 0

    2.

    public static intlastZero(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 (inti= 0; i< x.length; i++)
    {
    if (x[i] == 0)
    {
    return i;
    }
    } return -1;
    }
    // test: x=[0, 1, 0]
    // Expected = 2

    (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.

    解答:

    1.(1)故障为:循环中i的取值应当为i>=0

     (2)x=[2,3,5]  y=5

     (3)x=[2,3,5]  y=1

     (4)x=[2,3,5]  y=2

    2.(1)故障为:应当从后往前进行循环,应改为for(int i=x.length-1;i>=0;i--)

     (2)x=[1,2,3]

     (3)x=[1,0,2]

     (4)x=[0,1,0]

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  • 原文地址:https://www.cnblogs.com/wr344970835/p/5258429.html
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