Software Testing (ii)

Second assignment

For the following two programs

1. Find out fault

2, if possible, find out a test case does not execute fault

3, if possible, find a test case execution fault, but do not generate error

4. If possible, find a test case that produces an error, but does not cause failure

Public Intfindlast (int[] x, inty) {//effects:if x==null throw nullpointerexception//else return the index of the last El ement//in X is 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

　　

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    (int i= 0; i< x.length; i++)    {        if (x[i] = = 0)        {            ret Urn i;        }    } return-1;} Test:x=[0, 1, 0]//expected = 2

For the first paragraph of code

(1), for loop cannot traverse each element in the array, the control condition in the for loop should be modified to I >= 0

(2), if the x array is empty, it throws a null pointer and does not execute the fault

(3), test:x = [3,5,2]; y = 2;

Expected = 2

The first element in the array is not the element that needs to be found, so no error is generated

(4), test:x = [2,5,2]; y = 3;

Expected =-1

There is no element to be found in the array, so it returns-1, producing an error, but not generating failure.

For the second paragraph of code

(1), unable to find the last 0 position in the array as required, but to find out the position of the first 0 appears.

(2), if the x array is empty, it throws a null pointer and does not execute the fault

(3), test:x = [0,1,2];

Expected = 0

There is only one 0 in the array, no matter how it is found, it returns the 0 index, executes the fault, but does not produce an error

(4), test:x = [1,2,0,3];

Expected = 2

　　　　

Software Testing (ii)