通过秒数计算日期与时间

通过传入从START_YEAR到当前时间的秒数,计算当前日期与时间。

#define START_YEAR (1985)
#define SECOND_DAY    (86400)    //60*60*24
#define SECOND_HOUR    (3600)    //60*60
#define SECOND_MIN    (60)    //60


const unsigned short int mon_yday[][13] =
{
    /* Normal years.  */
    { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365 },
    /* Leap years.  */
    { 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366 }
};

//判断一个年份是否为闰年,是就返回1,不是就返回0
inline int isLeapYear(int year)
{
    return( (year%4 == 0 && year%100 != 0) || (year%400 == 0) );
}

//获取一年的天数
inline int getDaysForYear(int year)
{
    return (isLeapYear(year)?366:365);
}

//根据秒数计算日期
void getDate(int second, int& year, int& month, int& day)
{
    int days = second / SECOND_DAY;
    int curYear = START_YEAR;
    int leftDays = days;

    //calc year
    int daysCurYear = getDaysForYear(curYear);
    while (leftDays >= daysCurYear)
    {
        leftDays -= daysCurYear;
        curYear++;
        daysCurYear = getDaysForYear(curYear);
    }
    year = curYear;

    //calc month and day
    int isLeepYear = isLeapYear(curYear);
    for (int i = 1; i < 13; i++)
    {
        if (leftDays < mon_yday[isLeepYear][i])
        {
            month = i;
            day = leftDays - mon_yday[isLeepYear][i-1] + 1;
            break;
        }
    }
}

//计算时间
void getTime(int seconds, int& hour, int& minute, int& second)
{
    int leftSeconds = seconds % SECOND_DAY;
    hour = leftSeconds / SECOND_HOUR;
    minute = (leftSeconds % SECOND_HOUR) / SECOND_MIN;
    second = leftSeconds % SECOND_MIN;
}

原文地址:https://www.cnblogs.com/liuyunfeifei/p/3080199.html