Embed generic new() constraint into interface/class?

Question

So far I've been adding new() constraint like this.

public T Request<T>(BaseRequest request)
        where T : BaseResponse, new()

Is there a way to bake the new() into BaseResponse so that I only need to specify

        where T : BaseResponse

Answer 1

No. A derived class can always have a private/protected constructor, and there's no way to prevent that. It might even be an abstract class.

The only exception are struct s, which always have a default constructor - the one you get with default(SomeStruct) . Needless to say, this simply means the struct is initialized with zeroes (though C# 6 made this a bit more complicated).

Answer 2

The only way you could spare some typing, provided you have several generic methods in your class, is to make the class generic instead, ie go from this:

class YourClass
{
    public static T Request<T>(BaseRequest request)
       where T : BaseResponse, new()
    { }

    public static T SomethingElse<T>(BaseRequest request)
       where T : BaseResponse, new()
    { }
}

to this:

class YourClass<T> where T : BaseResponse, new()
{
    public static T Request(BaseRequest request)
    { }

    public static T SomethingElse(BaseRequest request)
    { }
}