How to do template specialization in C#

Question

In C#, the closest to specialization is to use a more-specific overload; however, this is brittle, and doesn't cover every possible usage. For example:

void Foo<T>(T value) {Console.WriteLine("General method");}
void Foo(Bar value) {Console.WriteLine("Specialized method");}

Here, if the compiler knows the types at compile, it will pick the most specific:

Bar bar = new Bar();
Foo(bar); // uses the specialized method

However....

void Test<TSomething>(TSomething value) {
    Foo(value);
}

will use Foo<T> even for TSomething=Bar, as this is burned in at compile-time.

One other approach is to use type-testing within a generic method - however, this is usually a poor idea, and isn't recommended.

Basically, C# just doesn't want you to work with specializations, except for polymorphism:

class SomeBase { public virtual void Foo() {...}}
class Bar : SomeBase { public override void Foo() {...}}

Here Bar.Foo will always resolve to the correct override.

Assuming you're talking about template specialization as it can be done with C++ templates - a feature like this isn't really available in C#. This is because C# generics aren't processed during the compilation and are more a feature of the runtime.

However, you can achieve similar effect using C# 3.0 extension methods. Here is an example that shows how to add extension method only for MyClass<int> type, which is just like template specialization. Note however, that you can't use this to hide default implementation of the method, because C# compiler always prefers standard methods to extension methods:

class MyClass<T> {
  public int Foo { get { return 10; } }
}
static class MyClassSpecialization {
  public static int Bar(this MyClass<int> cls) {
    return cls.Foo + 20;
  }
}

Now you can write this:

var cls = new MyClass<int>();
cls.Bar();

If you want to have a default case for the method that would be used when no specialization is provided, than I believe writing one generic Bar extension method should do the trick:

  public static int Bar<T>(this MyClass<T> cls) {
    return cls.Foo + 42;
  }

By adding an intermediate class and a dictionary, specialization is possible.

To specialize on T, we create an generic interface, having a method called (e.g.) Apply. For the specific classes that interface is implemented, defining the method Apply specific for that class. This intermediate class is called the traits class.

That traits class can be specified as a parameter in the call of the generic method, which then (of course) always takes the right implementation.

Instead of specifying it manually, the traits class can also be stored in a global IDictionary<System.Type, object>. It can then be looked up and voila, you have real specialization there.

If convenient you can expose it in an extension method.

class MyClass<T>
{
    public string Foo() { return "MyClass"; }
}

interface BaseTraits<T>
{
    string Apply(T cls);
}

class IntTraits : BaseTraits<MyClass<int>>
{
    public string Apply(MyClass<int> cls)
    {
        return cls.Foo() + " i";
    }
}

class DoubleTraits : BaseTraits<MyClass<double>>
{
    public string Apply(MyClass<double> cls)
    {
        return cls.Foo() + " d";
    }
}

// Somewhere in a (static) class:
public static IDictionary<Type, object> register;
register = new Dictionary<Type, object>();
register[typeof(MyClass<int>)] = new IntTraits();
register[typeof(MyClass<double>)] = new DoubleTraits();

public static string Bar<T>(this T obj)
{
    BaseTraits<T> traits = register[typeof(T)] as BaseTraits<T>;
    return traits.Apply(obj);
}

var cls1 = new MyClass<int>();
var cls2 = new MyClass<double>();

string id = cls1.Bar();
string dd = cls2.Bar();

See this link to my recent blog and the follow ups for an extensive description and samples.

I was searching for a pattern to simulate template specialization, too. There are some approaches which may work in some circumstances. However what about the case

static void Add<T>(T value1, T value2)
{
    //add the 2 numeric values
}

It would be possible to choose the action using statements e.g. if (typeof(T) == typeof(int)). But there is a better way to simulate real template specialization with the overhead of a single virtual function call:

public interface IMath<T>
{
    T Add(T value1, T value2);
}

public class Math<T> : IMath<T>
{
    public static readonly IMath<T> P = Math.P as IMath<T> ?? new Math<T>();

    //default implementation
    T IMath<T>.Add(T value1, T value2)
    {
        throw new NotSupportedException();    
    }
}

class Math : IMath<int>, IMath<double>
{
    public static Math P = new Math();

    //specialized for int
    int IMath<int>.Add(int value1, int value2)
    {
        return value1 + value2;
    }

    //specialized for double
    double IMath<double>.Add(double value1, double value2)
    {
        return value1 + value2;
    }
}

Now we can write, without having to know the type in advance:

static T Add<T>(T value1, T value2)
{
    return Math<T>.P.Add(value1, value2);
}

private static void Main(string[] args)
{
    var result1 = Add(1, 2);
    var result2 = Add(1.5, 2.5);

    return;
}

If the specialization should not only be called for the implemented types, but also derived types, one could use an In parameter for the interface. However, in this case the return types of the methods cannot be of the generic type T any more.