本文主要是介绍日期计算函数：已知当前日期，求过了一段时间后的日期。，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
函数的功能是，根据当前日期，计算过了一段时间后的日期。
1. 流程图：
2.函数实现：
/*** Method called to check if the date is valid. * short duration: unit is day. * short[] curData: current date, format is YYMMDD* byte [] expData: store the expected date, format is YYMMDD* Return value: false if fail, else succeed.*/private static boolean vGetExpectDate(short[] curDate, short duration, short [] expDate){short year;short month;short day;short days_left;short day_add = 0;// Assign the real date to year, month, day and houryear = curDate[0];month = curDate[1];day = curDate[2];day_add = duration;// days add to yearwhile(day_add > (days_left = sDaysToNextYear(year, month, day))){// come to the 1st day in 1st month of the next year.year++;month = 1;day = 1;day_add -= days_left;}// days add to monthwhile(day_add > (days_left = sDaysToNextMonth(year, month, day))){month++;day = 1;day_add -= days_left;}// days add to dayday += day_add;// store the expected date into the workbuf.expDate[0] = year;expDate[1] = month;expDate[2] = day;return true;}private static short sDaysToNextYear(byte year, byte month, byte day){byte days_Feb;byte days_month;short ret;switch(month){case 1:case 3:case 5:case 7:case 8:case 10:case 12:days_month = 31;break;case 4:case 6:case 9:case 11:days_month = 30;break;case 2:// Here assumes the year is 20xx, so if the year is leap or not depends on the last byte xx.if(0 == (year % 4))days_month = 29;elsedays_month = 28;break;default:break;}ret = days_month - day + 1;month++;switch(month){case 2:ret += days_Feb;case 3:ret += 31;case 4:ret += 30;case 5:ret += 31;case 6:ret += 30;case 7:ret += 31;case 8:ret += 31;case 9:ret += 30;case 10:ret += 31;case 11:ret += 30;case 12:ret += 31;default:break;}return ret;}private static short sDaysToNextMonth(byte year, byte month, byte day){byte days_month;switch(month){case 1:days_month = 31;break;case 2:// Here assumes the year is 20xx, so if the year is leap or not depends on the last byte xx.if(0 == (year % 4))days_month = 29;elsedays_month = 28;break;case 3:days_month = 31;break;case 4:days_month = 30;break;case 5:days_month = 31;break;case 6:days_month = 30;break;case 7:days_month = 31;break;case 8:days_month = 31;break;case 9:days_month = 30;break;case 10:days_month = 31;break;case 11:days_month = 30;break;case 12:days_month = 31;break;default:break;}return (days_month - day + 1);}这篇关于日期计算函数：已知当前日期，求过了一段时间后的日期。的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
