Visual Studio 2010/2012/2013,类图:如何将接口显示为基类,而不是"lillypop"？

Question

由于接口已经在图表上,我想明确地显示继承引用.但是我找不到......

Answer 1

VS 2005到2012年有一个错误,它不允许它工作.我有一个工作arround可能会欺骗它绘制接口的继承.假设您的界面名为IMyInterface.您必须将其替换为实现该接口的抽象类,并使用它而不是您的接口.代码将使用条件编译,如下所示:

//to generate class diagram, add 'CLSDIAGRAM' to the conditional symbols on the Build tab,
// or add '#define CLSDIAGRAM' at the top of this file
#if CLSDIAGRAM
#warning  CLSDIAGRAM is defined and this build should be used only in the context of class diagram generation
//rename your interface by adding _
public interface IMyInterface_ 
{
    int MyProperty { get; }
    void MyMethod();
}
//this class will act as an interface in the class diagram ;)
public abstract class IMyInterface : IMyInterface_ // tricks other code into using the class instead
{
//fake implementation
    public int MyProperty    {
        get { throw new NotImplementedException(); }
    }

    public void MyMethod()
    {
        throw new NotImplementedException();
    }
}
#else
// this is the original interface
public interface IMyInterface {
    int MyProperty { get; }
    void MyMethod();
}
#endif

这可能会按照您的意愿显示出来.在您的情况下,IMyInterface将成为IMedicine.