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.