这不是一个直接的答案，但更多的是从准科学的角度对当前问题进行哲学推理。

我认为，这个问题并没有具体说明衡量年龄的单位或文化，大多数答案似乎都假设了一个整数年表示。时间的国际单位制单位是秒，因此正确的通用答案应该是（当然，假设标准化日期时间，不考虑相对论效应）：

var lifeInSeconds = (DateTime.Now.Ticks - then.Ticks)/TickFactor;

在基督教以年计算年龄的方法中：

var then = ... // Then, in this case the birthday
var now = DateTime.UtcNow;
int age = now.Year - then.Year;
if (now.AddYears(-age) < then) age--;

在金融领域，当计算通常被称为日计数分数（Day Count Fraction）的东西时，也存在类似的问题，该分数大致是给定时期的年数。年龄问题确实是一个衡量时间的问题。

实际/实际（正确计算所有天数）惯例示例：

DateTime start, end = .... // Whatever, assume start is before end

double startYearContribution = 1 - (double) start.DayOfYear / (double) (DateTime.IsLeapYear(start.Year) ? 366 : 365);
double endYearContribution = (double)end.DayOfYear / (double)(DateTime.IsLeapYear(end.Year) ? 366 : 365);
double middleContribution = (double) (end.Year - start.Year - 1);

double DCF = startYearContribution + endYearContribution + middleContribution;

另一种很常见的衡量时间的方法通常是“序列化”（命名这一日期惯例的家伙一定是认真的“trippin”）：

DateTime start, end = .... // Whatever, assume start is before end
int days = (end - start).Days;

我想知道，在相对论年龄（以秒为单位）变得比迄今为止地球围绕太阳周期的粗略近似更有用之前，我们还需要多长时间：）或者换句话说，当一个周期必须给定一个位置或一个表示其自身运动的函数才能有效时：）

我对DateTime一无所知，但我能做的就是：

using System;
                    
public class Program
{
    public static int getAge(int month, int day, int year) {
        DateTime today = DateTime.Today;
        int currentDay = today.Day;
        int currentYear = today.Year;
        int currentMonth = today.Month;
        int age = 0;
        if (currentMonth < month) {
            age -= 1;
        } else if (currentMonth == month) {
            if (currentDay < day) {
                age -= 1;
            }
        }
        currentYear -= year;
        age += currentYear;
        return age;
    }
    public static void Main()
    {
        int ageInYears = getAge(8, 10, 2007);
        Console.WriteLine(ageInYears);
    }
}

有点困惑，但仔细看代码，这一切都是有意义的。

对此的简单答案是应用AddYears，如下所示，因为这是唯一一种将年份添加到闰年2月29日的本地方法，并获得普通年份2月28日的正确结果。

有些人认为3月1日是勒普林斯的生日，但.Net和任何官方规则都不支持这一点，也没有常见的逻辑解释为什么一些出生在2月的人应该在另一个月拥有75%的生日。

此外，Age方法可以作为DateTime的扩展添加。由此，您可以以最简单的方式获得年龄：

列表项目

int age=出生日期.age（）；

public static class DateTimeExtensions
{
    /// <summary>
    /// Calculates the age in years of the current System.DateTime object today.
    /// </summary>
    /// <param name="birthDate">The date of birth</param>
    /// <returns>Age in years today. 0 is returned for a future date of birth.</returns>
    public static int Age(this DateTime birthDate)
    {
        return Age(birthDate, DateTime.Today);
    }

    /// <summary>
    /// Calculates the age in years of the current System.DateTime object on a later date.
    /// </summary>
    /// <param name="birthDate">The date of birth</param>
    /// <param name="laterDate">The date on which to calculate the age.</param>
    /// <returns>Age in years on a later day. 0 is returned as minimum.</returns>
    public static int Age(this DateTime birthDate, DateTime laterDate)
    {
        int age;
        age = laterDate.Year - birthDate.Year;

        if (age > 0)
        {
            age -= Convert.ToInt32(laterDate.Date < birthDate.Date.AddYears(age));
        }
        else
        {
            age = 0;
        }

        return age;
    }
}

现在，运行此测试：

class Program
{
    static void Main(string[] args)
    {
        RunTest();
    }

    private static void RunTest()
    {
        DateTime birthDate = new DateTime(2000, 2, 28);
        DateTime laterDate = new DateTime(2011, 2, 27);
        string iso = "yyyy-MM-dd";

        for (int i = 0; i < 3; i++)
        {
            for (int j = 0; j < 3; j++)
            {
                Console.WriteLine("Birth date: " + birthDate.AddDays(i).ToString(iso) + "  Later date: " + laterDate.AddDays(j).ToString(iso) + "  Age: " + birthDate.AddDays(i).Age(laterDate.AddDays(j)).ToString());
            }
        }

        Console.ReadKey();
    }
}

关键日期示例如下：

出生日期：2000-02-29出生日期：2011-02-28年龄：11

输出：

{
    Birth date: 2000-02-28  Later date: 2011-02-27  Age: 10
    Birth date: 2000-02-28  Later date: 2011-02-28  Age: 11
    Birth date: 2000-02-28  Later date: 2011-03-01  Age: 11
    Birth date: 2000-02-29  Later date: 2011-02-27  Age: 10
    Birth date: 2000-02-29  Later date: 2011-02-28  Age: 11
    Birth date: 2000-02-29  Later date: 2011-03-01  Age: 11
    Birth date: 2000-03-01  Later date: 2011-02-27  Age: 10
    Birth date: 2000-03-01  Later date: 2011-02-28  Age: 10
    Birth date: 2000-03-01  Later date: 2011-03-01  Age: 11
}

2012年2月28日晚些时候：

{
    Birth date: 2000-02-28  Later date: 2012-02-28  Age: 12
    Birth date: 2000-02-28  Later date: 2012-02-29  Age: 12
    Birth date: 2000-02-28  Later date: 2012-03-01  Age: 12
    Birth date: 2000-02-29  Later date: 2012-02-28  Age: 11
    Birth date: 2000-02-29  Later date: 2012-02-29  Age: 12
    Birth date: 2000-02-29  Later date: 2012-03-01  Age: 12
    Birth date: 2000-03-01  Later date: 2012-02-28  Age: 11
    Birth date: 2000-03-01  Later date: 2012-02-29  Age: 11
    Birth date: 2000-03-01  Later date: 2012-03-01  Age: 12
}

我想添加希伯来文日历计算（或其他系统。全球化日历可以以相同的方式使用），使用此线程中的重写函数：

Public Shared Function CalculateAge(BirthDate As DateTime) As Integer
    Dim HebCal As New System.Globalization.HebrewCalendar ()
    Dim now = DateTime.Now()
    Dim iAge = HebCal.GetYear(now) - HebCal.GetYear(BirthDate)
    Dim iNowMonth = HebCal.GetMonth(now), iBirthMonth = HebCal.GetMonth(BirthDate)
    If iNowMonth < iBirthMonth Or (iNowMonth = iBirthMonth AndAlso HebCal.GetDayOfMonth(now) < HebCal.GetDayOfMonth(BirthDate)) Then iAge -= 1
    Return iAge
End Function

下面是一个测试片段：

DateTime bDay = new DateTime(2000, 2, 29);
DateTime now = new DateTime(2009, 2, 28);
MessageBox.Show(string.Format("Test {0} {1} {2}",
                CalculateAgeWrong1(bDay, now),      // outputs 9
                CalculateAgeWrong2(bDay, now),      // outputs 9
                CalculateAgeCorrect(bDay, now),     // outputs 8
                CalculateAgeCorrect2(bDay, now)));  // outputs 8

这里有一些方法：

public int CalculateAgeWrong1(DateTime birthDate, DateTime now)
{
    return new DateTime(now.Subtract(birthDate).Ticks).Year - 1;
}

public int CalculateAgeWrong2(DateTime birthDate, DateTime now)
{
    int age = now.Year - birthDate.Year;

    if (now < birthDate.AddYears(age))
        age--;

    return age;
}

public int CalculateAgeCorrect(DateTime birthDate, DateTime now)
{
    int age = now.Year - birthDate.Year;

    if (now.Month < birthDate.Month || (now.Month == birthDate.Month && now.Day < birthDate.Day))
        age--;

    return age;
}

public int CalculateAgeCorrect2(DateTime birthDate, DateTime now)
{
    int age = now.Year - birthDate.Year;

    // For leap years we need this
    if (birthDate > now.AddYears(-age)) 
        age--;
    // Don't use:
    // if (birthDate.AddYears(age) > now) 
    //     age--;

    return age;
}

还有一个答案：

public static int AgeInYears(DateTime birthday, DateTime today)
{
    return ((today.Year - birthday.Year) * 372 + (today.Month - birthday.Month) * 31 + (today.Day - birthday.Day)) / 372;
}

这已经过广泛的单元测试。它看起来确实有点“神奇”。数字372是如果每个月有31天，一年中会有多少天。

其工作原理的解释（此处省略）如下：

让我们设置Yn=DateTime.Now.Year，Yb=生日.Year，Mn=DateTime.Now.Month，Mb=生日.Month、Dn=DateTime.Now.Day，Db=生日.Day年龄=Yn-Yb+（31*（Mn-Mb）+（Dn-Db））/372我们知道，如果日期已经到达，我们需要的是Yn-Yb，如果日期尚未到达，则需要Yn-Yb-1。a） 如果Mn<Mb，我们有-341<=31*（Mn-Mb）<=-31和-30<=Dn-Db<=30-371<=31*（锰-Mb）+（Dn-Db）<=-1带整数除法（31*（Mn-Mb）+（Dn-Db））/372=-1b） 如果Mn=Mb和Dn<Db，则我们有31*（Mn-Mb）=0和-30<=Dn Db<=-1再次使用整数除法（31*（Mn-Mb）+（Dn-Db））/372=-1c） 如果Mn>Mb，我们有31<=31*（Mn-Mb）<=341和-30<=Dn-Db<=301<=31*（Mn-Mb）+（Dn-Db）<=371带整数除法（31*（Mn-Mb）+（Dn-Db））/372=0d） 如果Mn=Mb且Dn>Db，则我们有31*（Mn-Mb）=0且1<=Dn Db<=30再次使用整数除法（31*（Mn-Mb）+（Dn-Db））/372=0e） 如果Mn=Mb，Dn=Db，我们有31*（Mn-Mb）+Dn Db=0因此（31*（Mn-Mb）+（Dn-Db））/372=0