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软件测试二

时间:2018-03-14 22:33:28      阅读:241      评论:0      收藏:0      [点我收藏+]

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. Faultfor循环中应为i>=0

 

  1. Not executex=null ; y=3

 

  1. Not result in an error state: x=[3,3,3]; y=3

 

  1. Results in an error, but not a failurex=[0,0,0]; y=3

 

 

 

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. Fault:  for循环应为for(int i=x.length-1;i>=0;i--)

 

  1. Not executex=null ; y=3

 

  1. Not result in an error state:  x=[3]

 

  1. Results in an error, but not a failurex=[1,1,0]

 

 

 

软件测试二

原文:https://www.cnblogs.com/c-czl123/p/8570541.html

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