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：）

我使用ScArcher2的解决方案来精确计算人的年龄，但我需要进一步计算他们的月和日以及年。

    public static Dictionary<string,int> CurrentAgeInYearsMonthsDays(DateTime? ndtBirthDate, DateTime? ndtReferralDate)
    {
        //----------------------------------------------------------------------
        // Can't determine age if we don't have a dates.
        //----------------------------------------------------------------------
        if (ndtBirthDate == null) return null;
        if (ndtReferralDate == null) return null;

        DateTime dtBirthDate = Convert.ToDateTime(ndtBirthDate);
        DateTime dtReferralDate = Convert.ToDateTime(ndtReferralDate);

        //----------------------------------------------------------------------
        // Create our Variables
        //----------------------------------------------------------------------
        Dictionary<string, int> dYMD = new Dictionary<string,int>();
        int iNowDate, iBirthDate, iYears, iMonths, iDays;
        string sDif = "";

        //----------------------------------------------------------------------
        // Store off current date/time and DOB into local variables
        //---------------------------------------------------------------------- 
        iNowDate = int.Parse(dtReferralDate.ToString("yyyyMMdd"));
        iBirthDate = int.Parse(dtBirthDate.ToString("yyyyMMdd"));

        //----------------------------------------------------------------------
        // Calculate Years
        //----------------------------------------------------------------------
        sDif = (iNowDate - iBirthDate).ToString();
        iYears = int.Parse(sDif.Substring(0, sDif.Length - 4));

        //----------------------------------------------------------------------
        // Store Years in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Years", iYears);

        //----------------------------------------------------------------------
        // Calculate Months
        //----------------------------------------------------------------------
        if (dtBirthDate.Month > dtReferralDate.Month)
            iMonths = 12 - dtBirthDate.Month + dtReferralDate.Month - 1;
        else
            iMonths = dtBirthDate.Month - dtReferralDate.Month;

        //----------------------------------------------------------------------
        // Store Months in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Months", iMonths);

        //----------------------------------------------------------------------
        // Calculate Remaining Days
        //----------------------------------------------------------------------
        if (dtBirthDate.Day > dtReferralDate.Day)
            //Logic: Figure out the days in month previous to the current month, or the admitted month.
            //       Subtract the birthday from the total days which will give us how many days the person has lived since their birthdate day the previous month.
            //       then take the referral date and simply add the number of days the person has lived this month.

            //If referral date is january, we need to go back to the following year's December to get the days in that month.
            if (dtReferralDate.Month == 1)
                iDays = DateTime.DaysInMonth(dtReferralDate.Year - 1, 12) - dtBirthDate.Day + dtReferralDate.Day;       
            else
                iDays = DateTime.DaysInMonth(dtReferralDate.Year, dtReferralDate.Month - 1) - dtBirthDate.Day + dtReferralDate.Day;       
        else
            iDays = dtReferralDate.Day - dtBirthDate.Day;             

        //----------------------------------------------------------------------
        // Store Days in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Days", iDays);

        return dYMD;
}

我对Mark Soen的答案做了一个小小的修改：我重写了第三行，以便可以更容易地解析表达式。

public int AgeInYears(DateTime bday)
{
    DateTime now = DateTime.Today;
    int age = now.Year - bday.Year;            
    if (bday.AddYears(age) > now) 
        age--;
    return age;
}

为了清晰起见，我还将其转换为函数。

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);
}

以下方法（从.NET类DateDiff的时间段库中提取）考虑区域性信息的日历：

// ----------------------------------------------------------------------
private static int YearDiff( DateTime date1, DateTime date2 )
{
  return YearDiff( date1, date2, DateTimeFormatInfo.CurrentInfo.Calendar );
} // YearDiff

// ----------------------------------------------------------------------
private static int YearDiff( DateTime date1, DateTime date2, Calendar calendar )
{
  if ( date1.Equals( date2 ) )
  {
    return 0;
  }

  int year1 = calendar.GetYear( date1 );
  int month1 = calendar.GetMonth( date1 );
  int year2 = calendar.GetYear( date2 );
  int month2 = calendar.GetMonth( date2 );

  // find the the day to compare
  int compareDay = date2.Day;
  int compareDaysPerMonth = calendar.GetDaysInMonth( year1, month1 );
  if ( compareDay > compareDaysPerMonth )
  {
    compareDay = compareDaysPerMonth;
  }

  // build the compare date
  DateTime compareDate = new DateTime( year1, month2, compareDay,
    date2.Hour, date2.Minute, date2.Second, date2.Millisecond );
  if ( date2 > date1 )
  {
    if ( compareDate < date1 )
    {
      compareDate = compareDate.AddYears( 1 );
    }
  }
  else
  {
    if ( compareDate > date1 )
    {
      compareDate = compareDate.AddYears( -1 );
    }
  }
  return year2 - calendar.GetYear( compareDate );
} // YearDiff

用法：

// ----------------------------------------------------------------------
public void CalculateAgeSamples()
{
  PrintAge( new DateTime( 2000, 02, 29 ), new DateTime( 2009, 02, 28 ) );
  // > Birthdate=29.02.2000, Age at 28.02.2009 is 8 years
  PrintAge( new DateTime( 2000, 02, 29 ), new DateTime( 2012, 02, 28 ) );
  // > Birthdate=29.02.2000, Age at 28.02.2012 is 11 years
} // CalculateAgeSamples

// ----------------------------------------------------------------------
public void PrintAge( DateTime birthDate, DateTime moment )
{
  Console.WriteLine( "Birthdate={0:d}, Age at {1:d} is {2} years", birthDate, moment, YearDiff( birthDate, moment ) );
} // PrintAge

这是一种奇怪的方法，但如果您将日期设置为yyyymmdd，并从当前日期中减去出生日期，然后删除您获得的年龄的最后4位数字：）

我不知道C#，但我相信这在任何语言中都适用。

20080814 - 19800703 = 280111 

删除最后4位=28。

C#代码：

int now = int.Parse(DateTime.Now.ToString("yyyyMMdd"));
int dob = int.Parse(dateOfBirth.ToString("yyyyMMdd"));
int age = (now - dob) / 10000;

或者，也可以不进行扩展方法形式的所有类型转换。忽略错误检查：

public static Int32 GetAge(this DateTime dateOfBirth)
{
    var today = DateTime.Today;

    var a = (today.Year * 100 + today.Month) * 100 + today.Day;
    var b = (dateOfBirth.Year * 100 + dateOfBirth.Month) * 100 + dateOfBirth.Day;

    return (a - b) / 10000;
}