代碼段1

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

q1: Identify the fault
a1: 循環(huán)條件中i>0應(yīng)該改為i>=0, 如果不更改會忽略數(shù)組第一個數(shù)字
q2: If possible, identify a test case that does not execute the fault. (Reachability)
a2: test：x=[],y = 5
q3:f possible, identify a test case that executes the fault, but does not result in an error state.
a3: test: x=[2,3,5],y = 3;
q4:If possible identify a test case that results in an error, but not a failure.
a4: test: x=[2,3,5],y = 1;
代碼段2

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

q1: Identify the fault
a1: 循環(huán)應(yīng)該從數(shù)組末位開始，而原代碼中循環(huán)從數(shù)組首位開始，這樣找到的不是最后一個0而是第一個0
q2: If possible, identify a test case that does not execute the fault. (Reachability)
a2: x = []
q3:f possible, identify a test case that executes the fault, but does not result in an error state.
a3: x = [0]
q4:If possible identify a test case that results in an error, but not a failure.
a4: x = [1]

轉(zhuǎn)載于:https://www.cnblogs.com/IzayoiNamida/p/8570500.html

總結(jié)

以上是生活随笔為你收集整理的软件测试 homework2的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。