将Unix时间戳转换为YYYY-MM-DD HH :MM:SS

我有一个Unix时间戳，我需要从中获取各个年，月，日，小时，分钟和秒的值。我在数学课上从来都不是很出色，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。

最佳答案

免责声明:以下代码不解释leap years或leap seconds [Unix时间不解释leap秒。无论如何，他们被高估了。 -Ed]。另外，我没有进行测试，因此可能存在错误。它可能会踢您的猫并侮辱您的母亲。祝你今天过得愉快。

让我们尝试一些伪代码(Python，实际上):

# 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)

如果我发呆了，请告诉我。在C中执行此操作应该不会太困难。