阅读日期并打印出第二天日期的C程序

Question

嗨我刚刚开始在大学学习C语言。现在我面临着一个我不知道如何解决的问题。到目前为止，我们只使用了 库，没有别的，scanf也是我们下周要学的 。现在我们只是使用printf来处理所有事情。 我已经学会如何说如果一年是闰年，但是，我的任务是下一个：需要创建一个程序来读取日期并打印出第二天的日期，输出如下：

Enter a date in the form day month year: 17 5 2010 
The date of the next day is: 18/5/2010 

我的困境是，我不知道要使用什么操作或如何设置代码以确保考虑闰年，例如如果今天的日期是28 2 2010年下一个日期需要是1 3 2010年，因为它不是闰年。 使用的唯一库并没有scanf函数，但（与scanf函数还） 到目前为止，我得到这个：

#include <stdio.h> 

int day, month, year, ndays, leapdays; 
bool leapyear; 
int main() { 
    day = 28; 
    month = 2; 
    year = 2010; 
    ndays = day + 1; 
    leapdays = 31; 
    leapyear = false; 
    if (leapyear % 4 == 0) { 
     leapyear = true; 
    } 
    if (leapyear % 100 == 0) { 
     leapyear = false; 
    } 
    if (leapyear % 400 == 0) { 
     leapyear = true; 
    } 
    if ((leapyear) && (month == 12 || month == 1 || month == 3 || month == 5 
     || month == 7 || month == 8 || month == 10)) { 
     leapdays = 31; 
    } 
    if ((leapyear) && (month == 4 || month == 6 || month == 9 || month == 11 
     )) { 
     leapdays = 30; 
    } 
    if ((leapyear) && (month == 2)) { 
     leapdays = 29; 
    } else if ((leapyear == false) && (month == 2)) { 
     leapdays = 28; 
    } 
    printf ("Enter a date in the form day month year: %d %d %d \n", day, 
      month, year); 
    printf ("The date of the next day is: %d/%d/%d", ndays, month, year); 
} 

Answer 1

你加1天之后，检查，看看是否能值比数较大当月的天数。如果是这样，请将日期设置为1并将月份加1。然后检查月份是否大于12，如果是，则将月份设置为1并将年份加1。

至于确定一个月中的天数，除了二月份以外的所有月份都具有相同的天数，无论该年份是否为闰年。现在你正在检查年份是否是其他月份的闰年。您可以退出该检查并只检查月份编号。

if (month == 12 || month == 1 || month == 3 || month == 5 
    || month == 7 || month == 8 || month == 10) { 
    leapdays = 31; 
} 
if (month == 4 || month == 6 || month == 9 || month == 11) { 
    leapdays = 30; 
} 

Answer 2

明白了吧:)感谢所有帮助

#include <stdio.h> 

    int day, month, year, ndays, leapdays; 
    bool leapyear; 
    int main() { 
     day = 31; 
     month = 12; 
     year = 2010; 
     ndays = day + 1; 
     leapyear = false; 
     printf ("Enter a date in the form day month year: %d %d %d \n", day, month, year); 
     if (year % 4 == 0) { 
      leapyear = true; 
     } 
     if (year % 100 == 0) { 
      leapyear = false; 
     } 
     if (year % 400 == 0) { 
      leapyear = true; 
     } 
     if ((leapyear) && (month == 12 || month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10)) { 
      leapdays = 31; 
     } 
     if ((leapyear) && (month == 4 || month == 6 || month == 9 || month == 11)) { 
      leapdays = 30; 
     } 
     if ((leapyear) && (month == 2)) { 
      leapdays = 29; 
     } else if (leapyear == false) { 
      leapdays = 28; 
     } 
     if ((leapdays == 31) && (day == 31)) { 
      ndays = 1; 
      month = month + 1; 
     }else if ((leapdays == 30) && (day == 30)) { 
        ndays = 1; 
        month = month + 1; 

     }else if ((leapdays == 29) && (day == 29)) { 
         ndays = 1; 
        month = month + 1; 
     }else if ((leapdays == 28) && (day == 28)) { 
        ndays = 1; 
        month = month + 1; 
     }else if ((month == 12) && (day == 31)) { 
        ndays = 1; 
        month = 1; 
        year = year + 1; 
     } 


     printf ("The date of the next day is: %d/%d/%d", ndays, month, year); 
    } 

Answer 3

考虑不同的流量。首先先找到每月的日子，然后再测试月底和年底。

int year, month, day; 
    // set year, month, day in some fashion 

    day++; // tomorrow 

    int days_per_month = 31; 
    if (month == 4 || month == 6 || month == 9 || month == 11) { 
    days_per_month = 30; 
    } else if (month == 2) { 
    days_per_month = 28; 
    if (year % 4 == 0) { 
     days_per_month = 29; 
     if (year % 100 == 0) { 
     days_per_month = 28; 
     if (year % 400 == 0) { 
      days_per_month = 29; 
     } 
     } 
    } 
    } 
    if (day > days_per_month) { 
    day = 1; 
    month++; 
    if (month > 12) { 
     month = 1; 
     year++; 
    } 
    } 

其他改善将使用辅助功能，枚举类型和不同的定义。然而这段代码似乎反映了OP的水平。