homework2
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一、findLast()
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
(a) The for-loop should include 0:
for (int i=x.length-1; i>=0; i--)
(b) The null value for x will result in a NullPointerException before the loop test is evaluated hence no execution of the fault.
Input: x = null; y=3
Expected Output: NullPointerException
Actual Output: NullPointerException
(c) For any input where y appears in the second or later position, there is no error. Also, if x is empty, there is no error.
Input: x=[2,3,5]; y=3;
Expected Output: 1
Acutal Output: 1
(d) For an input where y is not in x, the missing path(i.e. An incorrect PC on the final loop that is not taken) is an error, but there is no failure.
Input: x=[2,3,5]; y=7;
Expected Output: -1
Actual Output: -1
二、lastZero()
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
(a)The for-loop should search high to low:
for (int i=x.length-1; i>=0; i--)
(b)All inputs execute the fault, even the null input.
(c)If the loop is not unrolled at all, there’s no error.Also if the loop is only unrolled once, high-to-low and low-to-high evaluation are the same. Hence there is no error for length 0 or length 1 inputs.
Input: x = [3]
Expected Output: -1
Actual Output: -1
(d)There’s an error anytime the loop is unrolled more than once, since the values of index i ascend instead of descend.
Input: x = [1,0,3]
Expected Output: 1
Actual Output: 1
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