Question one:Identify the fault.
the answer:
the first function :The for-loop misses the 0 index,so that should be -- for(int i=x.length-1;i>=0;i--)
the second function: The for-loop is from beginning to the end ,but we want to find the last 0. So that should be -- for(int i=x.length-1;i>=0;i--)
Question two:If possible, identify a test case that does not execute the fault. (Reachability)
The answer:
the first function:x==null will throw a NullPointerException and it doesn't enter the for-loop test ,so there is no fault to execute. Just like this: Input: x=null;y=0 Expected:NullPointerException The real output: NullPointerException
the second function:there is not such a case.
Question three:If possible, identify a test case that executes the fault, but does not result in an error state.
The answer:
the first function: the y doesn't appear in the first position,there is no error. Just like this: Input: x=[0,2,0];y= 2; Expected:1 The real output: 1
the second function: if the for-loop just executes once, that should be ok. Because the first become the last. Just like this: Input: x=[0] Expected:1 The real output: 1
Question four:If possible identify a test case that results in an error, but not a failure.
The answer:
the first function: the y doesn't appear in the x array. Just like this: Input: x=[0,2,0];y= 3; Expected:-1 The real output: -1
the second function: if the for-loop just executes more than once and the 0 appears only once, that should be ok. Just like this: Input: x=[1,0,2] Expected:1 The real output: 1