Exactly. Option<T> has a implicit conversion operator that converts a type of T to Option<T>. So passing T to the function constructs a new Option<T>, and the

Exactly. Option<T> has a implicit conversion operator that converts a type of T to Option<T>. So passing T to the function constructs a new Option<T>, and the constructor checks for null. It's also likely (though not guaranteed) that the Option<T> will be allocated on the stack, and should therefore not pose too much of an overhead.