case1:
1. Identify the fault.
for循环中的条件判断应为:(int i=x.length-1; i > =0;i–);
2. If possible, identify a test case that does not execute the fault. (Reachability)
test: x=[];
(抛出空指针异常,没有执行下面的程序,则没有执行fault)
3. If possible, identify a test case that executes the fault, but does not result in an error state.
test: x=[3, 2, 5]; y = 2
Expected= 1
(执行了含有fault的程序,但是并没有产生错误,即执行了fault,没有执行error)
4. If possible identify a test case that results in an error, but not a failure.
test: x=[3, 2, 5]; y = 1 Expected = -1
(没有遍历x=1,直接返回了-1,因此执行了error,没有执行failure。)
case2:
1、 Identify the fault.
for循环中的条件判断应为:(int i=x.length-1; i > =0; i–);
2、 If possible, identify a test case that does not execute the fault. (Reachability)
test: x=[];
(抛出空指针异常,没有执行下面的程序,则没有执行fault)
3、 If possible, identify a test case that executes the fault, but does not result in an error state.
test: x=[1, 2, 0]
Expected = 2
(执行了含有fault的程序,但是并没有产生错误,即执行了fault,没有执行error)
4、 If possible identify a test case that results in an error, but not a failure.
test: x=[3, 2, 5];
Expected = -1
(遍历到i=3时,返回了-1,不符合设计目的,因此执行了error,没有执行failure。)
posted @
2018-03-15 22:19
齐某人qg
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