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
Identify the fault.
for循环中
i >0
修改为
i >= 0
If possible, identify a test case that does not execute the fault. (Reachability)
x = null, y = 0
If possible, identify a test case that executes the fault, but does not result in an error state.
x = [2,3,4] , y = 4
If possible identify a test case that results in an error, but not a failure.
x = [2,3,4] , y = 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
Identify the fault.
for循环中应该从后往前遍历修改为
for(int i = x.length-1; i >= 0; i--)
If possible, identify a test case that does not execute the fault. (Reachability)
x = null
If possible, identify a test case that executes the fault, but does not result in an error state.
x = [0]
If possible identify a test case that results in an error, but not a failure.
x = [0,1,2]
转载于:https://www.cnblogs.com/tjliqy/p/8570663.html