NSubstitute使用out参数模拟一个void方法

时间:2015-05-19 15:16:37

标签: c# unit-testing testing mocking nsubstitute

我是NSubstitute的新手,我试图使用2个void参数模拟out方法,我很确定我做错了。

我有一个CustomerDataAccess类,其方法具有以下签名:

void GetCustomerWithAddresses(int customerId, 
                              out List<Customer> customers, 
                              out List<Address> addresses);

CustomerRepository调用其GetCustomer方法,然后调用CustomerDataAccess.GetCustomerWithAddresses DAL方法。然后,DAL方法输出两个out参数,一个用于客户,一个用于地址。然后,存储库方法使用AutoMapper将DAL方法中的两个对象映射到存储库随后返回的业务域。

这是我到目前为止的代码,它不起作用。我的研究并没有帮助我确定解决这个问题需要做些什么。如何设置out参数的值?

// Arange
ICustomerDataAccess customerDataAccess = Substitute.For<ICustomerDataAccess>();
IList<Customer> customers;
IList<Address> addresses;

customerDataAccess.When(x => x.GetCustomerWithAddresses(
    1, out customers, out addresses))
    .Do(x =>
    {
        customers = new List<Customer>() { new Customer() { CustomerId = 1, CustomerName = "John Doe" } };
        addresses = new List<Address>() { new Address() { AddressId = 1, AddressLine1 = "123 Main St", City = "Atlanta" } };
    });

CustomerRepository sut = new CustomerRepository(customerDataAccess);

// Act
Customer customer = sut.GetCustomer(1);

// Assert
Assert.IsNotNull(customer);

2 个答案:

答案 0 :(得分:16)

out参数使用其参数位置作为索引进行更新。它在Returns documentation for NSubstitute中进行了解释。因此,对于您的特定情况,您正在填充第二个和第三个参数,因此您应该像这样设置您的呼叫:

customerDataAccess.When(x => x.GetCustomerWithAddresses(1, out customers, out addresses))
.Do(x =>
{
    x[1] = new List<Customer>() { new Customer() { CustomerId = 1, CustomerName = "John Doe" } };
    x[2] = new List<Address>() { new Address() { AddressId = 1, AddressLine1 = "123 Main St", City = "Atlanta" } };
});

答案 1 :(得分:0)

对于非void方法,可以使用常规返回语法:

 var haveWithAddresses = customerDataAccess.GetCustomerWithAddresses(1, out customers, out addresses)
               .Returns(callInfo => { 
                     callInfo[0] = new List<Customer>();
                     callInfo[1] = new List<Address>();
                     return true;
               });

使用Void方法的When...Do语法是正确的。