I want to create a generic class that has a member of type T. T may be a class, a nullable class, a struct, or a nullable struct. So basically anything. This is a simplified example that shows my problem:
#nullable enable
class Box<T> {
public T Value { get; }
public Box(T value) {
Value = value;
}
public static Box<T> CreateDefault()
=> new Box<T>(default(T));
}
Due to using the new #nullable enable feature I get the following warning: Program.cs(11,23): warning CS8653: A default expression introduces a null value when 'T' is a non-nullable reference type.
This warning makes sense to me. I then tried to fix it by adding a ? to the property and constructor parameter:
#nullable enable
class Box<T> {
public T? Value { get; }
public Box(T? value) {
Value = value;
}
public static Box<T> CreateDefault()
=> new Box<T>(default(T));
}
But now I get two errors instead:
Program.cs(4,12): error CS8627: A nullable type parameter must be known to be a value type or non-nullable reference type. Consider adding a 'class', 'struct', or type constraint.
Program.cs(6,16): error CS8627: A nullable type parameter must be known to be a value type or non-nullable reference type. Consider adding a 'class', 'struct', or type constraint.
However, I don't want to add a constraint. I don't care if T is a class or a struct.
An obvious solution is to wrap the offending members under a #nullable disable directive. However, like #pragma warning disable, I'd like to avoid doing that unless it's necessary. Is there another way in getting my code to compile without disabling the nullability checks or the CS8653 warning?
$ dotnet --info
.NET Core SDK (reflecting any global.json):
Version: 3.0.100-preview4-011223
Commit: 118dd862c8