使用 GetElementType() 时获取空引用

Getting a null reference while using GetElementType()

我正在尝试使我的 graphQL 突变变得通用。但是在运行时我得到一个错误。
错误提示:未将对象引用设置为对象的实例。

public class ApplicationMutation<T> : ObjectGraphType where T: BaseModel
{
    public ApplicationMutation()
    {
        this.Name = "RootMutation";

        var nameofT = typeof(T).GetElementType().Name;

        this.Field<AddPayloadType<T>>(
            "add" + nameofT,
            arguments: new InputQueryArguments<AddInputType<T>>(),
            resolve: context =>
            {
                var input = context.GetArgument<AddInput<T>>("input");

                var result = Activator.CreateInstance<T>();
                {
                    Name = "1337 p40c355I73m";
                };

                return new AddPayload<T>(input, result);
            });

        this.Field<UpdatePayloadType<T>>(
            "update" + nameofT,
            arguments: new InputQueryArguments<UpdateInputType<T>>(),
            resolve: context =>
            {
                var input = context.GetArgument<UpdateInput<T>>("input");

                var result = Activator.CreateInstance<T>();
                {
                    Name = "rul0r item";
                };

                return new UpdatePayload<T>(input, result);
            });

        this.Field<DeletePayloadType<T>>(
            "delete" + nameofT,
            arguments: new InputQueryArguments<DeleteInputType<T>>(),
            resolve: context =>
            {
                var input = context.GetArgument<DeleteInput<T>>("input");

                var result = true;

                return new DeletePayload<T>(input, result);
            });
    }
}

从以下行抛出异常: var nameofT = typeof(T).GetElementType().Name;

如果需要更多信息,请提出要求。

您只能将 GetElementType() 用于数组、指针和引用类型..

null if the current Type is not an array or a pointer, or is not passed by reference, or represents a generic type or a type parameter in the definition of a generic type or generic method https://msdn.microsoft.com/en-us/library/system.type.getelementtype.aspx

它将 return null 用于仿制药