如何比较C＆＃x2B;＆＃x2B中的两个DateTime结构有效？

Question

我有以下DateTime结构：

struct DateTime
{
    std::uint16_t year;
    std::uint8_t month;
    std::uint8_t day;
    std::uint8_t hour;
    std::uint8_t minute;
    std::uint8_t second;
    std::uint16_t milisecond;
};

我的疑问是关于 Lessthan 和更大的方法方法。我的实施如下，以避免一堆 ifs 和 elses ，但我可能没有涵盖所有可能的情况：

bool GreaterThan(const DateTime& datetime)
{
    bool greater{true};

    // When found a different value for the most significant value, the evaluation is interrupted
    if ((year <= datetime.year) && (month <= datetime.month || year < datetime.year) &&
        (day <= datetime.day || month < datetime.month) && (hour <= datetime.hour || day < datetime.day) &&
        (minute <= datetime.minute || hour < datetime.hour) &&
        (second <= datetime.second || minute < datetime.minute) &&
        (milisecond <= datetime.milisecond || second < datetime.second))
    {
        greater = false;
    }

    return greater;
}

bool LessThan(const DateTime& datetime)
{
    bool less{true};

    // When found a different value for the most significant value, the evaluation is interrupted
    if ((year >= datetime.year) && (month >= datetime.month || year > datetime.year) &&
        (day >= datetime.day || month > datetime.month) && (hour >= datetime.hour || day > datetime.day) &&
        (minute >= datetime.minute || hour > datetime.hour) &&
        (second >= datetime.second || minute > datetime.minute) &&
        (milisecond >= datetime.milisecond || second > datetime.second))
    {
        less = false;
    }

    return less;
}

请让我知道哪种情况不是覆盖。

Answer 1

从所有评论中，我认为最好的解决方案是由 -
Richard Critten ：

bool GreaterThan(const DateTime& date_time)
{
    return (std::tie(year, month, day, hour, minute, second, milisecond) > 
            std::tie(date_time.year, date_time.month, date_time.day, date_time.hour, date_time.minute, date_time.second, date_time.milisecond));
}

bool LessThan(const DateTime& date_time)
{
    return (std::tie(year, month, day, hour, minute, second, milisecond) <
            std::tie(date_time.year, date_time.month, date_time.day, date_time.hour, date_time.minute, date_time.second, date_time.milisecond));
}

Answer 2

您的实施看起来过于痛苦。有一种更简单，合乎逻辑的方法来实现这一目标。如果您知道贪婪的算法，那是与此类似的思考过程。

bool GreaterThan(const datetime& datetime)
{
    if(year != datetime.year) return year > datetime.year;
    if(month != datetime.month) return month > datetime.month;
    //... and so on (omitted)
    return false; // they are the same
}

您可以类似地实现Lessthan。