将 unix 时间戳转换为 YYYY-MM-DD HH :MM:SS

Question

我有一个 Unix 时间戳，我需要从中获取单独的年、月、日、小时、分钟和秒值。我的数学课一直不太好，所以我想知道你们是否可以帮我一点忙 :)

我必须自己做所有事情(没有 time.h 函数)。语言是C。

Answer 1

免责声明:以下代码不考虑leap years <罢工>或leap seconds [Unix 时间不考虑闰秒。无论如何，他们被高估了。 -Ed]。另外，我没有测试它，所以可能有错误。它可能会踢你的猫并侮辱你的母亲。祝你有美好的一天。

让我们尝试一些伪代码(实际上是 Python):

# Define some constants here...

# You'll have to figure these out.  Don't forget about February (leap years)...
secondsPerMonth = [ SECONDS_IN_JANUARY, SECONDS_IN_FEBRUARY, ... ]

def formatTime(secondsSinceEpoch):
    # / is integer division in this case.
    # Account for leap years when you get around to it :)
    year = 1970 + secondsSinceEpoch / SECONDS_IN_YEAR
    acc = secondsSinceEpoch - year * SECONDS_IN_YEAR

    for month in range(12):
        if secondsPerMonth[month] < acc:
            acc -= month
            month += 1

    month += 1

    # Again, / is integer division.
    days = acc / SECONDS_PER_DAY
    acc -= days * SECONDS_PER_DAY

    hours = acc / SECONDS_PER_HOUR
    acc -= hours * SECONDS_PER_HOUR

    minutes = acc / SECONDS_PER_MINUTE
    acc -= minutes * SECONDS_PER_MINUTE

    seconds = acc

    return "%d-%d-%d %d:%d%d" % (year, month, day, hours, minutes, seconds)

如果我搞砸了，请告诉我。在 C 中执行此操作应该不会太难。