这个解决方案怎么样？

static string CalcAge(DateTime birthDay)
{
    DateTime currentDate = DateTime.Now;         
    int approximateAge = currentDate.Year - birthDay.Year;
    int daysToNextBirthDay = (birthDay.Month * 30 + birthDay.Day) - 
        (currentDate.Month * 30 + currentDate.Day) ;

    if (approximateAge == 0 || approximateAge == 1)
    {                
        int month =  Math.Abs(daysToNextBirthDay / 30);
        int days = Math.Abs(daysToNextBirthDay % 30);

        if (month == 0)
            return "Your age is: " + daysToNextBirthDay + " days";

        return "Your age is: " + month + " months and " + days + " days"; ;
    }

    if (daysToNextBirthDay > 0)
        return "Your age is: " + --approximateAge + " Years";

    return "Your age is: " + approximateAge + " Years"; ;
}

==常见说法（从几个月到几岁）===

如果您只是为了通用，以下是代码作为您的信息：

DateTime today = DateTime.Today;
DateTime bday = DateTime.Parse("2016-2-14");
int age = today.Year - bday.Year;
var unit = "";

if (bday > today.AddYears(-age))
{
    age--;
}
if (age == 0)   // Under one year old
{
    age = today.Month - bday.Month;

    age = age <= 0 ? (12 + age) : age;  // The next year before birthday

    age = today.Day - bday.Day >= 0 ? age : --age;  // Before the birthday.day

    unit = "month";
}
else {
    unit = "year";
}

if (age > 1)
{
    unit = unit + "s";
}

测试结果如下：

The birthday: 2016-2-14

2016-2-15 =>  age=0, unit=month;
2016-5-13 =>  age=2, unit=months;
2016-5-14 =>  age=3, unit=months; 
2016-6-13 =>  age=3, unit=months; 
2016-6-15 =>  age=4, unit=months; 
2017-1-13 =>  age=10, unit=months; 
2017-1-14 =>  age=11, unit=months; 
2017-2-13 =>  age=11, unit=months; 
2017-2-14 =>  age=1, unit=year; 
2017-2-15 =>  age=1, unit=year; 
2017-3-13 =>  age=1, unit=year;
2018-1-13 =>  age=1, unit=year; 
2018-1-14 =>  age=1, unit=year; 
2018-2-13 =>  age=1, unit=year; 
2018-2-14 =>  age=2, unit=years; 

这里有一个单行线：

int age = new DateTime(DateTime.Now.Subtract(birthday).Ticks).Year-1;

通过较少的转换和UtcNow，这段代码可以照顾闰年2月29日出生的人：

public int GetAge(DateTime DateOfBirth)
{
    var Now = DateTime.UtcNow;
    return Now.Year - DateOfBirth.Year -
        (
            (
                Now.Month > DateOfBirth.Month ||
                (Now.Month == DateOfBirth.Month && Now.Day >= DateOfBirth.Day)
            ) ? 0 : 1
        );
}

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);
}

这里有一个非常简单且易于遵循的示例。

private int CalculateAge()
{
//get birthdate
   DateTime dtBirth = Convert.ToDateTime(BirthDatePicker.Value);
   int byear = dtBirth.Year;
   int bmonth = dtBirth.Month;
   int bday = dtBirth.Day;
   DateTime dtToday = DateTime.Now;
   int tYear = dtToday.Year;
   int tmonth = dtToday.Month;
   int tday = dtToday.Day;
   int age = tYear - byear;
   if (bmonth < tmonth)
       age--;
   else if (bmonth == tmonth && bday>tday)
   {
       age--;
   }
return age;
}