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Moq-如何设置一个懒惰的界面

[英]Moq - How to setup a Lazy interface

I want to mock a Lazy Interface and also Setup a method to return false. 我想模拟一个惰性接口,还要Setup一个返回false的方法。

The problem is, when i run the test, I get a NotSupportedException: 问题是,当我运行测试时,出现NotSupportedException:

System.NotSupportedException: 'Invalid setup on a non-virtual (overridable in VB) member: mock => mock.Value System.NotSupportedException:'在非虚拟(在VB中可重写)成员上的无效设置:模拟=>模拟值

Here is a simplified example: 这是一个简化的示例:

[TestMethod]
public void SomeTestMethod()
{
    var someService = new Mock<Lazy<IService>>();

    /*this line throws the exception*/  
    someService.Setup(x => x.Value.SomeMethod()).Returns(false);
    ...
}

Please consider that SomeMethod is actually virtual, but somehow getting the lazy initialization using x.Value is not supported by Moq. 请考虑SomeMethod实际上是虚拟的,但是Moq不支持使用x.Value进行延迟初始化。

I didn't found a solution for this specific scenario but I did view some other approaches on declarations, but sadly didn't work for me. 我没有找到针对这种特定情况的解决方案,但我确实查看了声明的其他方法,但可惜对我没有用。

[TestMethod]
public void SomeTestMethod()
{
    var someService = new Mock<IService>();
    var lazySomeService = new Lazy<IService>(() => someService.Object);

    //tried this but won't compile
    //lazySomeService.Setup(x => x.Value.SomeMethod()).Returns(false);
    //lazySomeService.Value.Setup(x => x.SomeMethod()).Returns(false);
    ...
}

You started on the right track with 您从正确的轨道开始

var someService = new Mock<IService>();
var lazySomeService = new Lazy<IService>(() => someService.Object);

but the setup needs to be on the mock not the actual Lazy implementation. 但是设置需要模拟而不是实际的Lazy实现。

someService.Setup(x => x.SomeMethod()).Returns(false);

That way when the Lazy.Value is called, it will be using the mock. 这样,当调用Lazy.Value时,它将使用模拟。

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