Question

I need to make sure that a datatype implements the IComparable interface, and I was wondering if there was anyway to make that a requirement when you create the object?

Answer 1

You can perhaps use generic to do this, for example:

public static T Create<T>() where T : IComparable<T>, new() {     return new T(); } 

Or if you mean ‘when you create the type’ (in code), then no; you’d just have to remember, perhaps using unit-tests to verify that you’ve done it.

I recommend using the typed IComparable<T> over IComparable – it makes life a lot easier (and avoids some boxing, but that is less of an issue). Finally, remember that you can use Comparer<T>.Default and Comparer.Default in code if you want to duck-type the comparable bit (like how List<T>.Sort() works).