如何在返回模拟对象之前修改方法调用的结果？

Question

鉴于以下简化示例，使用RhinoMocks和MSpec：

[Subject(typeof (LocationController))]
public class when_creating_a_location_with_invalid_model : context_for_location_controller
{
    static LocationModel model = new LocationModel();
    static SelectList states = new SelectList(new Dictionary<string,string> {
        { "IN", "Indiana" }, { "NY", "New York" }
    });

    static ActionResult result;

    Establish context = () =>
        {
            LocationModelBuilder.Stub(x =>
                x.Build(Arg<LocationModel>.Is.Equal(model))).Return(model);
        }

    Because of = () => result = subject.Create(model);

    It should_automatically_select_a_state = () => result.Model<LocationModel>()
         .States.ShouldNotBeEmpty();
}

如何从 LocationModelBuilder.Build（）的存根调用返回之前修改 model 变量中包含的对象？我想在返回 Build（）之前执行model.States = states之类的任务。我尝试使用 Do（）处理程序，但我放弃了......

Answer 1

尝试使用WhenCalled（）。 WhenCalled参数允许访问模拟方法的参数，您也可以设置返回值。

.WhenCalled(m => {
   Model model = (Model) m.Arguments[0];
   model.States = ...;
});