视频

(2010/3/11 修改成 youku)

c/c++求两个日期之间的间隔天数/** //z 2014-04-22 13:19:45 L.253 BG57IV3@XCL T1903383406.K.F253293061 [T269,L3801,R176,V5040]参见msdn tm time_t 注意有效范围，里面的year不能太早，否则计算不准确 */  int day_distance_1(const int year1,const int month1,const int day1,const int year2,const int month2,const int day2)  {      struct tm tm1;      tm1.tm_year = year1 - 1900;      tm1.tm_mon = month1 - 1;      tm1.tm_mday = day1;      tm1.tm_hour = 0;      tm1.tm_min = 0;      tm1.tm_sec = 0;        struct tm tm2;      tm2.tm_year = year2 - 1900;      tm2.tm_mon = month2 - 1;      tm2.tm_mday = day2;      tm2.tm_hour = 0;      tm2.tm_min = 0;      tm2.tm_sec = 0;        time_t time1;      time_t time2;      time1 = mktime(&tm1);      time2 = mktime(&tm2);      double diff = difftime(time1,time2);      return (int)(diff/(3600*24));  }  /** 这个方法的计算范围很大*/  int day_distance_2(const int year1,const int month1,const int day1,const int year2,const int month2,const int day2)  {      int nd, nm, ny; //new_day, new_month, new_year      int od, om, oy; //old_day, oldmonth, old_year          nm = (month2 + 9) % 12;      ny = year2 - nm/10;      nd = 365*ny + ny/4 - ny/100 + ny/400 + (nm*306 + 5)/10 + (day2 - 1);          om = (month1 + 9) % 12;      oy = year1 - om/10;      od = 365*oy + oy/4 - oy/100 + oy/400 + (om*306 + 5)/10 + (day1 - 1);          return od - nd;  }  int main()  {      cout << day_distance_1(2012,1,14,2011,9,21) << endl;      cout << day_distance_2(2012,1,14,2011,9,21) << endl;  }//z 2014-04-22 13:19:45 L.253 BG57IV3@XCL T1903383406.K.F253293061 [T269,L3801,R176,V5040]