使类依赖项出现在接口中而不是属性

发布于 2025-01-08 04:31:35 字数 431 浏览 0 评论 0原文

我有一个接口和一个实现该接口的类。

我要求该类依赖于 ITimer。创建私有成员并传入构造函数参数，并将私有成员分配给该构造函数参数。

如何使接口以某种方式声明此 ITimer 但不作为属性？这并不是说我无法正确测试 MyClass，但我认为接口应该是类必须遵守的合同，因此我的类具有依赖项，那么我如何使其出现在接口中？

public interface IMyInterface
{
  void DoSomething();
}

public class MyClass : IMyInterface
{
  private ITimer MyTimer;

  public MyClass(ITimer timer)
  {
    MyTimer = timer;
  }

  public void DoSomething()
  {

  }
}

原文

I have an interface and a class that implements that interface.

I have a requirement that the class has a dependency on ITimer. A private member is created and a constructor argument is passed in and the private member is assigned to that constructor parameter.

How do I make it so that the interface has this ITimer declared somehow but not as a property? Its not as if I can't test MyClass properly but I thought the interface was supposed to be a contract that the class must adhere to therefore my class has a dependency so how do I make this appear in the interface?

public interface IMyInterface
{
  void DoSomething();
}

public class MyClass : IMyInterface
{
  private ITimer MyTimer;

  public MyClass(ITimer timer)
  {
    MyTimer = timer;
  }

  public void DoSomething()
  {

  }
}

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找回味觉 2025-01-15 04:31:35

这无法通过 C# 中的接口来完成。您可以做的是从抽象类继承而不是接口。

public abstract class AMyClass
{
  protected ITimer MyTimer;

  public AMyClass(ITimer timer)
  {
    MyTimer = timer;
  }

  public abstract void DoSomething();
}

public class MyClass : AMyClass
{
  public override void DoSomething()
  {

  }
}

This cannot be done with an interface in C#. What you could do is to inherit from an abstract class instread of an interface.

public abstract class AMyClass
{
  protected ITimer MyTimer;

  public AMyClass(ITimer timer)
  {
    MyTimer = timer;
  }

  public abstract void DoSomething();
}

public class MyClass : AMyClass
{
  public override void DoSomething()
  {

  }
}

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