rhino-mocks - 如何在模拟对象返回之前修改方法调用的结果？

Question

给定以下简化示例，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();
}

在从LocationModelBuilder.Build()的存根调用返回之前，如何修改模型变量中包含的对象？我想在Build()返回之前执行分配。我尝试使用Do()处理程序，但我放弃了......model.States = states

Answer 1

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

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