Moq First()Last()和GetEnumerator()怪异 [英] Moq First() Last() and GetEnumerator() wierdness

问题描述

我是Moqing，我的路线零件来自 rps = new List ... (3 Route Parts)

I am Moqing my Route Parts from a rps = new List... (3 Route Parts)

和Moqing GetEnumerator()如下所示

route.Setup(ro => ro.GetEnumerator()).Returns(rps.GetEnumerator());

，但是Moq在以下代码中失败，并且在调用Last()

but the Moq fails in the following code with "Sequence contains no elements" on the call to Last()

o.Route.Any(rp => rp.IsNonTowLocation && 
rp != o.Route.First() && 
rp != o.Route.Last())

在即时窗口中查看First() Last()，我发现如果多次执行First() Last()，则值会更改.好像是MoveNext()被调用但没有被Reset()调用，并且Enumerable完全混淆了.有没有人通过Moq遇到过这种情况并找到了解决方案?

Looking at First() Last() in the immediate windows I find the values change if I execute First() Last() multiple times. Its as if the MoveNext() gets called but not Reset() and the Enumerable it totally confused. Has anyone experienced this with Moq and found a solution?

推荐答案

您已经设置了GetEnumerator，以便每次都返回相同的枚举器实例.

You've setup your GetEnumerator so that it returns the same enumerator instance every time.

route.Setup(ro => ro.GetEnumerator()).Returns(rps.GetEnumerator());

这等效于:

var enumerator = rps.GetEnumerator()
route.Setup(ro => ro.GetEnumerator()).Returns(enumerator);

如果您希望在每次调用时都使用一个新的枚举数，则需要传递一个返回lambda表达式:

If you want a new enumerator on every call, then you need to pass Returns a lambda expression:

route.Setup(ro => ro.GetEnumerator()).Returns(() => rps.GetEnumerator());

每次调用GetEnumerator()时都会调用lambda-因此First()和Last()应该可以按预期工作.

The lambda will get called every time GetEnumerator() is called - so First() and Last() should then work as expected.

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