单项选择题
下列有关软件测试设计的说法中,正确的是______。
A) 测试方案应考虑是否可行、是否有效和是否能够达到预期的测试目标
B) 基于判定表的测试用例设计方法是白盒测试用例设计方法
C) 测试方案设计中可以忽略软件系统的实际使用环境
D) 测试开发不是测试用例设计的工作内容
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已知C源程序如下: *Input today’s date,output tomorrow’s date * * version 2 * #include<stdio. h> struct ydate int day; int month; int year;; int leap(struct ydate d) if((d. year%4==0&&d. year%100 ! =0)||(d. year%400==0)) return 1; else return 0; int numdays(struct ydate d) int day; static int daytab[]= 31,28,31,30,31,30,3l,31,30,31,30,31; if(1eap(d)&&d. month==2) day=29; else day=daytabEd. month-1]; return day; int main(void) struct ydate today,tomorrow;printf( format of date is:year,month,day输入的年、月、日之间应用逗号隔开 n); printf( today is: ); scanf(“%d,%d.% ,&today.year,&today.month,&today.day); while(0>=today. year || today. year>65535||0>=today. month||today. month>12)|| 0>=today. day||today. day>numdays(today)) printf( input date error!reenter the day! n ); printf( today is: ); scanf( %d,%d,%d ,&today. year,&today.month,&today. day); if(today. day!=numdays(today)) tomorrow. year=today. year; tomorrow. month=today. month; tomorrow. day=today. day+1; else if(today.month==12) tomorrow. year=today. year+1; tomorrow. month=1; tomorrow. day=1; else tomorrow. year=today. year; tomorrow. month=today. month+1: tomorrow.day=1; printf( tomorrow is:%d,%d,%d n n , tomorrow. year,tomorrow. month,tomorrow. day); (1) 画出程序中所有函数的控制流程图; (2) 设计一组测试用例,使该程序所有函数的语句覆盖率和分支覆盖率均能达到100%。如果认为该程序的语句或分支覆盖率无法达到100%,需说明为什么。
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请根据说明,划分等价类。
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使用等价类划分方法为该软件设计测试用例。