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Something just occurred to me earlier today that has got me scratching my head. Any variable of type Nullable<T> can be assigned to null . For instance: int? i = null; At first I couldn't see how this would be possible without somehow defining an implicit conversion from object to Nullable<T> : public static implicit operator Nullable<T>(object box); But the above operator clearly does not exist, as if it did then the following would also have to be legal, at least at compile-time (which it isn't): int? i = new object(); Then I realized that perhaps the Nullable<T> type could define an implicit conversion to some arbitrary reference type that can never be instantiated, like this: public abstract class DummyBox { private DummyBox() { } } public struct Nullable<T> where T : struct { public static implicit operator Nullable<T>(DummyBox box) { if (box == null) { return new Nullable<T>(); } // This should never be possible, as a DummyBox cannot be instantiated. throw new InvalidCastException(); } } However, this does not explain what occurred to me next: if the HasValue property is false for any Nullable<T> value, then that value will be boxed as null : int? i = new int?(); object x = i; // Now x is null. Furthermore, if HasValue is true , then the value will be boxed as a T rather than a T? : int? i = 5; object x = i; // Now x is a boxed int, NOT a boxed Nullable<int>. But this seems to imply that there is a custom implicit conversion from Nullable&

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