模拟惰性接口会导致 运行 次异常

Question

我在尝试正确模拟包含在 Lazy 中的任意接口时遇到困难 class。

我试过：

[TestClass]
public class MyFooServiceTests
{
    private Mock<Lazy<IFoo>> _lazyFooMock = new Mock<Lazy<IFoo>>();
    private Mock<IFoo> _fooMock = new Mock<IFoo>();
    private MyFooService _service;

    [TestMethod]
    public void FooMethod_HappyPath_ShouldReturn()
    {
        //Arrange
        _fooMock
            .Setup(x => x.DoSomething())
            .Returns(1);

        _lazyFooMock
            .SetupGet(x => x.Value)
            .Returns(_fooMock.Object); // --------> Throws Exception.

        _service = new MyService(_lazyFooMock.Object);
    }
}

public interface IFoo
{
    int DoSomething();
}

public class MyFooService
{
    public MyFooService(IFoo foo) { ... }
}

异常消息：

Unsupported expression: x => x.Value Non-overridable members (here: Lazy.get_Value) may not be used in setup / verification expressions.

起订量：4.16.1

Answer 1

在其他人的帮助下解决了，我放弃了 _lazyFooMock 并将其替换为实际的 Lazy：_lazyFoo，并在重载的帮助下启动了它：public Lazy(Func<T> valueFactory)。

private Mock<IFoo> _fooMock = new Mock<IFoo>();
private Lazy<IFoo> _lazyFoo;

private MyFooService _service;

[TestMethod]
public void FooMethod_HappyPath_ShouldReturn()
{
    //Arrange
    _fooMock
        .Setup(x => x.DoSomething())
        .Returns(1);

    _lazyFoo = new Lazy<IFoo>(() => _fooMock.Object);

    _service = new MyFooService(_lazyFoo);
}