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：）

2需要解决的主要问题有：

1.计算准确年龄-以年、月、日等为单位。

2.计算人们普遍认为的年龄——人们通常不关心自己到底多大，他们只关心自己当年的生日是什么时候。

1的解决方案显而易见：

DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today;     //we usually don't care about birth time
TimeSpan age = today - birth;        //.NET FCL should guarantee this as precise
double ageInDays = age.TotalDays;    //total number of days ... also precise
double daysInYear = 365.2425;        //statistical value for 400 years
double ageInYears = ageInDays / daysInYear;  //can be shifted ... not so precise

2的解决方案在确定总年龄时并不那么精确，但人们认为它是精确的。当人们“手动”计算年龄时，通常也会使用它：

DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today;
int age = today.Year - birth.Year;    //people perceive their age in years

if (today.Month < birth.Month ||
   ((today.Month == birth.Month) && (today.Day < birth.Day)))
{
  age--;  //birthday in current year not yet reached, we are 1 year younger ;)
          //+ no birthday for 29.2. guys ... sorry, just wrong date for birth
}

注释2.：

这是我的首选解决方案我们不能使用DateTime.DayOfYear或TimeSpans，因为它们会在闰年中改变天数为了可读性，我只增加了几行

还有一个提示。。。我将为它创建两个静态重载方法，一个用于通用，另一个用于使用友好：

public static int GetAge(DateTime bithDay, DateTime today) 
{ 
  //chosen solution method body
}

public static int GetAge(DateTime birthDay) 
{ 
  return GetAge(birthDay, DateTime.Now);
}

我已经创建了一个SQL Server用户定义函数来计算某人的年龄，给定他们的出生日期。当您需要它作为查询的一部分时，这很有用：

using System;
using System.Data;
using System.Data.Sql;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;

public partial class UserDefinedFunctions
{
    [SqlFunction(DataAccess = DataAccessKind.Read)]
    public static SqlInt32 CalculateAge(string strBirthDate)
    {
        DateTime dtBirthDate = new DateTime();
        dtBirthDate = Convert.ToDateTime(strBirthDate);
        DateTime dtToday = DateTime.Now;

        // get the difference in years
        int years = dtToday.Year - dtBirthDate.Year;

        // subtract another year if we're before the
        // birth day in the current year
        if (dtToday.Month < dtBirthDate.Month || (dtToday.Month == dtBirthDate.Month && dtToday.Day < dtBirthDate.Day))
            years=years-1;

        int intCustomerAge = years;
        return intCustomerAge;
    }
};

我认为TimeSpan包含了我们所需要的一切，而不必求助于365.25（或任何其他近似值）。扩展Aug的示例：

DateTime myBD = new DateTime(1980, 10, 10);
TimeSpan difference = DateTime.Now.Subtract(myBD);

textBox1.Text = difference.Years + " years " + difference.Months + " Months " + difference.Days + " days";

这为这个问题提供了“更多细节”。也许这就是你要找的

DateTime birth = new DateTime(1974, 8, 29);
DateTime today = DateTime.Now;
TimeSpan span = today - birth;
DateTime age = DateTime.MinValue + span;

// Make adjustment due to MinValue equalling 1/1/1
int years = age.Year - 1;
int months = age.Month - 1;
int days = age.Day - 1;

// Print out not only how many years old they are but give months and days as well
Console.Write("{0} years, {1} months, {2} days", years, months, days);

我创建了一个Age结构，如下所示：

public struct Age : IEquatable<Age>, IComparable<Age>
{
    private readonly int _years;
    private readonly int _months;
    private readonly int _days;

    public int Years  { get { return _years; } }
    public int Months { get { return _months; } }
    public int Days { get { return _days; } }

    public Age( int years, int months, int days ) : this()
    {
        _years = years;
        _months = months;
        _days = days;
    }

    public static Age CalculateAge( DateTime dateOfBirth, DateTime date )
    {
        // Here is some logic that ressembles Mike's solution, although it
        // also takes into account months & days.
        // Ommitted for brevity.
        return new Age (years, months, days);
    }

    // Ommited Equality, Comparable, GetHashCode, functionality for brevity.
}