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C語言小程序如何判斷兩個日期之差

更新時間：2013年07月22日 11:32:16 作者：

輸入兩個日期，計算之間相差多少天。用了兩種方法實現(xiàn)，第二種利用結(jié)構(gòu)體，代碼比較清晰，其余的都一樣

1.普通的寫法

#include <stdio.h>
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 < 0)?(-diff):diff;
  }
  else
  {
   if(day1[1] < day2[1])    //day1=1991-5-8   day2=1991-6-2
   {
    tmp = day1;      //day1=1991-6-2   day2=1991-5-8
    day1 = day2;
    day2 = tmp;
   }
   for(i=day2[1]+1; i<day1[1]; i++)
   {
    diff +=  month[i];
   }
   diff += month[day2[1]] - day2[2] + day1[2];
   if(day2[1] <= 2 && day1[1] >2)
    if(leapyear(day2[0]))
     diff++;
  }
 }
 else
 {
  if(day1[0] < day2[0])
  {
   tmp = day1; 
   day1 = day2;
   day2 = tmp;
  }
  for(i=day2[0]+1; i<day1[0]; i++)
  {
   if(leapyear(i))
    diff += 366;
   else
    diff += 365;
  }
  for(i=day2[1]+1; i<=12; i++)     //day1=1992-1-1   day2=1991-1-1
  {
   diff += month[i];
  }
  diff += (month[day2[1]] - day2[2]);
  if(day2[1] <= 2)
   if(leapyear(day2[0]))
    diff++;
  for(i=1; i<day1[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("輸入另一個日期：");
 scanf("%d-%d-%d",&day2[0], &day2[1], &day2[2]);
 day = days(day1, day2);
 printf("兩個日期之間共有%d天。\n",day);
 return 0;
}

2.利用結(jié)構(gòu)體，代碼更整潔一些

復(fù)制代碼代碼如下:

#include <stdio.h>
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)    //如果第一個日期比第二個日期大，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; i<date2->month; i++)
   {
    diff += month[i];
   }
   diff += month[date1->month] - date1->day + date2->day;
   if(leapyear(date1->year))
    if(date1->month <=2 && date2->month >2)
     diff++;
  }
 }
 else
 {
  for(i=date1->year+1; i<date2->year; i++)
  {
   if(leapyear(i))
    diff += 366;
   else 
    diff += 365;
  }
  for(i=date1->month+1; i<=12; i++)    //date1距離年末多少天
  {
   diff += month[i];
  }
  diff += month[date1->month] - date1->day;
  if(date1->month <= 2)
   if(leapyear(date1->year))
    diff++;
  for(i=1; i<date2->month; 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("輸入另一個日期：");
 scanf("%d-%d-%d",&date2->year, &date2->month, &date2->day);
 compare(date1, date2);
 days = diff(date1, date2);
 printf("兩個日期之間共有%d天。\n",days);
 return 0;
}