题目:
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[]=
{31,28,31,30,31,30,31,31,30,31,30,31);
if(leap(d)&&d.month==2)
day=29;
else
day=daytab[d.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,%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\",
tomorrow.year,tomorrow.month,tomorrow.day);
}
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。 |
答案:
参考答案:
解析:根据以上分析,设计测试用例如下:
用例 编号 | 年 | 月 | 日 | leap | numdays | while | if…else if …else | 输出结果 |
1 | 2008 | 2 | 30 | 1 | 29 | 1 | error | |
2 | 2007 | 12 | 31 | 0 | 31 | 0 | month=12 | 2008.1.1 |
3 | 6 | 10 | 31 | 0 | 31 | 0 | today= numdays | 6.11.1 |
4 | 804 | 4 | 17 | 1 | 30 | 0 | today!= numdays | 804.4.18 |