Question

我试图在C中与另一个约会一个日期，并在几天内找到它们之间的差异。然而，这比它最初出现的要复杂得多，因为我显然必须考虑到不同年份的不同日期，因为闰年和不同的天数取决于它是哪个月。我确实试过使用time.h但是我认为它不允许闰年。

目前，我的数据作为整数存储在数组中，例如{2010, 5, 1, 2011, 6, 1}。

那么有人可以发帖或指点我的算法来帮助我完成这项任务吗？非常感谢你。

Answer 1

标准库C Time Library包含您想要的结构和功能。

此头文件包含要获取和的函数的定义操纵日期和时间信息。

同时检查C date and time functions和C program days between two dates

修改： -

试试这个：

int main() { int day1,mon1,year1,day2,mon2,year2; int ref,dd1,dd2,i; clrscr(); printf("Enter first date day, month, year\n"); scanf("%d%d%d",&day1,&mon1,&year1); printf("Enter second date day, month, year\n"); scanf("%d%d%d",&day2,&mon2,&year2); ref = year1; if(year2<year1) ref = year2; dd1=0; dd1=dater(mon1); for(i=ref;i<year1;i++) { if(i%4==0) dd1+=1; } dd1=dd1+day1+(year1-ref)*365; dd2=0; for(i=ref;i<year2;i++) { if(i%4==0) dd2+=1; } dd2=dater(mon2)+dd2+day2+((year2-ref)*365); printf("\n\n Difference between the two dates is %d days",abs(dd2-dd1)); getch(); } int dater(x) { int y=0; switch(x) { case 1: y=0; break; case 2: y=31; break; case 3: y=59; break; case 4: y=90; break; case 5: y=120;break; case 6: y=151; break; case 7: y=181; break; case 8: y=212; break; case 9: y=243; break; case 10:y=273; break; case 11:y=304; break; case 12:y=334; break; default: printf("Invalid Input\n\n\n\n"); exit(1); } return(y); }

或使用time.h尝试这样：

#include <stdio.h> #include <time.h> int main () { struct tm start_date; struct tm end_date; time_t start_time, end_time; double seconds; start_date.tm_hour = 0; start_date.tm_min = 0; start_date.tm_sec = 0; start_date.tm_mon = 10; start_date.tm_mday = 15; start_date.tm_year = 113; end_date.tm_hour = 0; end_date.tm_min = 0; end_date.tm_sec = 0; end_date.tm_mon = 10; end_date.tm_mday = 20; end_date.tm_year = 113; start_time = mktime(&start_date); end_time = mktime(&end_date); seconds = difftime(end_time, start_time); printf ("%.f seconds difference\n", seconds); return 0; }

Answer 2

ALGO：

将日期转换为struct tm。
将struct tm转换为time_t。
取time_t之间的差异。它产生日期之间的差异，以秒为单位。
将差值除以86400，将其转换为天数。

Answer 3

/* This function calculates the differance between two dates, passes 6 parameters date-month-year of first date and date-month-year of second date and prints the difference in date-weeks-year  format */
void Date::subtract(int firstDate, int firstMonth, int firstYear,int secondDate, int secondMonth, int secondYear)
{
/*check the dates are valid or not  */
if(isDateValid(firstDate,firstMonth,firstYear) && isDateValid(secondDate,secondMonth,secondYear) )
{

    firstMonth = (firstMonth + 9) % 12;
    firstYear = firstYear - firstMonth / 10;
    FirstNoOfDays = 365 * firstYear + firstYear/4 - firstYear/100 + firstYear/400 + (firstMonth * 306 + 5) /10 + ( firstDate - 1 );

    secondMonth = (secondMonth + 9) % 12;
    secondYear = secondYear - secondMonth / 10;
    SecondNoOfDays = 365 * secondYear + secondYear/4 - secondYear/100 + secondYear/400 + (secondMonth * 306 + 5) /10 + ( secondDate - 1 );

    dayDifference = abs(FirstNoOfDays - SecondNoOfDays); /* uses absolute if the first date is smaller so it wont give negative number */
    years = dayDifference / 365;
    weeks = (dayDifference % 365)/7;
    days = (dayDifference % 365) % 7;
    cout<<years<<" years "<<weeks<<" weeks "<<days<< " days"<<endl;

}
else
{
    cout<<"Invalid Date"<<endl;
}

}

找出两个日期之间的差异C.

3 个答案: