Question

i 在试图利用演播室和企业建筑师汇编我的节目时,会发现这一错误。

I m 撰写企业建筑师工具,我必须做图表,我继续犯这一错误,即不知道做什么。

存在问题的法典是:

    public Graph(EA.Repository repository)
    {

        EA.Diagram maindiagram;
        this.modelRepository = repository;
        maindiagram = repository.GetCurrentDiagram(); //recupero del diagramma
        this.diagramId = maindiagram.DiagramID; //identificativo del diagramma

        //inizializzazione nodi
        Collection nodeCollection = maindiagram.DiagramObjects;
        nodeList = new ArrayList();


        foreach (DiagramObject diagram in maindiagram.DiagramObjects)
        {
            diagramList.Add(diagram);
            foreach (Element element in diagramList)
            {
                if (element.Type == "Class"|| element.Type == "Component"||element.Type == "Package")
                { nodeList.Add(new Node(diagram, ref repository)); }

            }                

        }

        //inizializzazione archi 
        Collection linkCollection = maindiagram.DiagramLinks;
        linkList = new ArrayList();

        foreach (DiagramLink edge in maindiagram.DiagramLinks)
        {
            edgeList.Add(edge);
            foreach(Connector connector in edgeList)
                if (connector.Type == "Association" || connector.Type == "Aggregation" || connector.Type == "Compose" || connector.Type == "Dependency" 
                    || connector.Type == "Generalization" || connector.Type == "Realization")
                { linkList.Add(new Link (edge, ref repository));}
        }

请了解情况。

谢谢大家!

Answer 1

It s impossible to understand what is going on in this code, just by message provided.

查阅特例电文提供的线数,并按行
或》和“例外情况”中破除“实际上关于产生问题的界线”。

Answer 2

repository could be null - you should be doing null checks on parameter arguments to determine whether you can proceed or not; also maindiagram could be null (for all we know, if repository is something, then GetCurrentDiagram might return null.

这两种情况都是以可能造成你问题的方式得到的。

这种方法的范围不大,因此假定其范围更为宽松:这也不例外,但你称之为<代码>。增加。您也认为,如果是这样的话,如果不检查<条码><> > > > > > > >null ,则试图进入its<>em> 性质。

简言之,在你颁布的法典中可能出现这种情况的地方很多。但是,答案是:,其中无。

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