C語(yǔ)言中怎樣判斷兩個(gè)日期之差？輸入兩個(gè)日期，計(jì)算之間相差多少天。 用了兩種方法實(shí)現(xiàn)，第二種利用結(jié)構(gòu)體，代碼比較清晰，其余的都一樣，下面來(lái)一起看看吧。
1.普通的寫(xiě)法
?
#include
int leapyear(int year)
{
?if((year%4==0 && year%100!=0) || year%400==0)
??return 1;
?else
??return 0;
}
int days(int *day1, int *day2)
{
?int i=0;
?int *tmp;
?int diff = 0;
?const int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
?if(day1[0] == day2[0])
?{
??if(day1[1] == day2[1])
??{
???diff = day1[2] - day2[2];
???diff = (diff ??}
??else
??{
???if(day1[1] ???{
????tmp = day1;??????//day1=1991-6-2?? day2=1991-5-8
????day1 = day2;
????day2 = tmp;
???}
???for(i=day2[1]+1; i???{
????diff +=? month[i];
???}
???diff += month[day2[1]] - day2[2] + day1[2];
???if(day2[1] 2)
????if(leapyear(day2[0]))
?????diff++;
??}
?}
?else
?{
??if(day1[0] ??{
???tmp = day1;?
???day1 = day2;
???day2 = tmp;
??}
??for(i=day2[0]+1; i??{
???if(leapyear(i))
????diff += 366;
???else
????diff += 365;
??}
??for(i=day2[1]+1; i??{
???diff += month[i];
??}
??diff += (month[day2[1]] - day2[2]);
??if(day2[1] ???if(leapyear(day2[0]))
????diff++;
??for(i=1; i??{
???diff += month[i];
??}
??diff += day1[2];
??if(day1[1] > 2)
???if(leapyear(day1[0]))
????diff++;
?}
?return diff;
}
int main()
{
?int day1[3], day2[3];
?int day = 0;
?printf("輸入日期：");
?scanf("%d-%d-%d",&day1[0], &day1[1], &day1[2]);
?printf("輸入另一個(gè)日期：");
?scanf("%d-%d-%d",&day2[0], &day2[1], &day2[2]);
?day = days(day1, day2);
?printf("兩個(gè)日期之間共有%d天。/n",day);
?return 0;
}
2.利用結(jié)構(gòu)體，代碼更整潔一些
?
?
#include
typedef struct date
{
?int year;
?int month;
?int day;
}DATE;
int leapyear(int year)
{
?if((year%4==0 && year%100!=0) || year%400==0)
??return 1;
?else
??return 0;
}
int compare(DATE *d1, DATE *d2)????//如果第一個(gè)日期比第二個(gè)日期大，交換日期
{
?DATE *tmp;
?if(d1->year == d2->year)????//年數(shù)相等
?{
??if(d1->month > d2->month)???//月數(shù)相等
??{
???tmp = d1;
???d1 = d2;
???d2 = d1;
??}
??else if(d1->month == d2->month)??//日期相等
??{
???if(d1->day > d2->day)
???{
????tmp = d1;
????d1 = d2;
????d2 = d1;
???}
??}
?}
?else if(d1->year > d2->year)
?{
??tmp = d1;
??d1 = d2;
??d2 = tmp;
?}
?return 0;
}
int diff(DATE *date1, DATE *date2)
{
?int i;
?int diff = 0;
?const int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
?if(date1->year == date2->year)
?{
??if(date1->month == date2->month)
??{
???diff = date2->day - date1->day;
??}
??else
??{
???for(i=date1->month+1; imonth; i++)
???{
????diff += month[i];
???}
???diff += month[date1->month] - date1->day + date2->day;
???if(leapyear(date1->year))
????if(date1->month month >2)
?????diff++;
??}
?}
?else
?{
??for(i=date1->year+1; iyear; i++)
??{
???if(leapyear(i))
????diff += 366;
???else
????diff += 365;
??}
??for(i=date1->month+1; i??{
???diff += month[i];
??}
??diff += month[date1->month] - date1->day;
??if(date1->month ???if(leapyear(date1->year))
????diff++;
??for(i=1; imonth; i++)?????//date2距離年初多少天
??{
???diff += month[i];
??}
??diff += date2->day;
??if(date1->month > 2)
???if(leapyear(date2->year))
????diff++;
?}
?return diff;
}
int main()
{
?int days = 0;
?DATE day1, day2;
?DATE *date1, *date2;
?date1 = &day1;
?date2 = &day2;
?printf("輸入日期：");
?scanf("%d-%d-%d",&(date1->year), &(date1->month), &(date1->day));
?printf("輸入另一個(gè)日期：");
?scanf("%d-%d-%d",&date2->year, &date2->month, &date2->day);
?compare(date1, date2);
?days = diff(date1, date2);
?printf("兩個(gè)日期之間共有%d天。/n",days);
?return 0;
}
小編帶來(lái)的這篇C語(yǔ)言中怎樣判斷兩個(gè)日期之差，是不是特別的實(shí)用呢，更多關(guān)于C語(yǔ)言的內(nèi)容，可以多多關(guān)注下武林技術(shù)頻道，我們會(huì)不斷的更新。
