C# – Interop type cannot be embedded

An easy to understand and simple solution.

// Save today's date.
var today = DateTime.Today;

// Calculate the age.
var age = today.Year - birthdate.Year;

// Go back to the year in which the person was born in case of a leap year
if (birthdate.Date > today.AddYears(-age)) age--;

However, this assumes you are looking for the western idea of the age and not using East Asian reckoning.

All are different.

• typeof takes a type name (which you specify at compile time).
• GetType gets the runtime type of an instance.
• is returns true if an instance is in the inheritance tree.

Example

class Animal { } 
class Dog : Animal { }

void PrintTypes(Animal a) { 
    Console.WriteLine(a.GetType() == typeof(Animal)); // false 
    Console.WriteLine(a is Animal);                   // true 
    Console.WriteLine(a.GetType() == typeof(Dog));    // true
    Console.WriteLine(a is Dog);                      // true 
}

Dog spot = new Dog(); 
PrintTypes(spot);

What about typeof(T)? Is it also resolved at compile time?

Yes. T is always what the type of the expression is. Remember, a generic method is basically a whole bunch of methods with the appropriate type. Example:

string Foo<T>(T parameter) { return typeof(T).Name; }

Animal probably_a_dog = new Dog();
Dog    definitely_a_dog = new Dog();

Foo(probably_a_dog); // this calls Foo<Animal> and returns "Animal"
Foo<Animal>(probably_a_dog); // this is exactly the same as above
Foo<Dog>(probably_a_dog); // !!! This will not compile. The parameter expects a Dog, you cannot pass in an Animal.

Foo(definitely_a_dog); // this calls Foo<Dog> and returns "Dog"
Foo<Dog>(definitely_a_dog); // this is exactly the same as above.
Foo<Animal>(definitely_a_dog); // this calls Foo<Animal> and returns "Animal". 
Foo((Animal)definitely_a_dog); // this does the same as above, returns "Animal"