创建流程图的算法[一点指导??



好吧,我知道这是一个模糊的问题,但我似乎在这里被逻辑所困......我想创建输入程序的流程图。我从两天开始就一直在考虑这个问题,但无法获得最佳的总体方法......所以我拼命地看着你们在这里帮我......可能是我错过了一些小东西....

我有一个xml文件,其中包含有关给定java程序的信息,如下所示:

<Method modifier="publicstatic" type="void" name="main" >
    <FormalParameter modifier="" type="String[]" var_name="args" />
    <Throw>NullPointerException</Throw>
    <Throw>
        IndexOutofBoundException
    </Throw>
    <Field modifier="" type="int" name="x,y,z" />
    <Field modifier="" type="int" name=" sum[] " />
    <If>
        condition><![CDATA[(x==0)]]></condition>
        <Statement>
            <![CDATA[System.out.Println("I am in true")]]></Statement>
        <If>
            <condition><![CDATA[(y==2)]]></condition>
            <Statement>
                <![CDATA[System.out.Println("I am in
    true of y==2")]]></Statement>
        </If>
        <Statement>
            <![CDATA[System.out.prnitln("I am in
    true again")]]></Statement>
    </If>
    <else>
        <If>
            <condition><![CDATA[(x==2)]]></condition>
            <Statement>
                <![CDATA[System.out.println("I am in
    x==2 true")]]></Statement>
        </If>
        <else>
            <Statement>
                <![CDATA[System.out.println("I am in
     else 2")]]></Statement>
        </else>
    </else>
    <Statement>
        <![CDATA[c=b+d]]></Statement>
    <Statement>
        <![CDATA[a=b+c]]></Statement>
</Method>

现在这是生成的 xml 文件的一部分。冷杉代码片段:

public static void main(String[] args) throws NullPointerException,IndexOutofBoundException {
    int x,y,z;
    int sum[]={1,2,3,4};
    if(x==0)
    {
        System.out.Println("I am in true");
        if(y==2)
        {
            System.out.Println("I am in true of y==2");
        }
        System.out.prnitln("I am in true again");
    }
    else if(x==2)
    {
        System.out.println("I am in x==2 true");
    }
    else
    {
        System.out.println("I am in else 2");
    }
    c=b+d;
    a=b+c;   
}

现在我的做法是:

我有一个读取xml文件的类阅读器和一个负责绘图的类createFLowChart。我开始遍历方法节点...如果我找到一个语句,我调用一个函数FundStatement,它绘制一个reangular box并将其与prv节点连接。对 If-else 构造执行此操作,而无需添加大量状态布尔变量。那么有人可以指导我吗?

现在问题在于 if-else 构造。我找不到一种简单的方法来遍历 if -else 树并在节点之间正确建立边缘连接,而无需添加许多变量来包含状态信息,即 if 已启动或启动等。这是我的方法,但我发现深度调整很困难,即连接嵌套的 if-else 语句的叶子:

public void traverse(String path,CreateFlowChart parent)
{
    xPath =  XPathFactory.newInstance().newXPath();
    XPathExpression expr;
    //ArrayList&lt;dataObjects.Interface&gt; parentInterfaces=new ArrayList&lt;dataObjects.Interface&gt;();
    //dataObjects.Class[] classes=new dataObjects.Class[90];
    try {
        expr = xPath.compile(path);
        Object result = expr.evaluate(document, XPathConstants.NODESET);
        NodeList MethodNodes = (NodeList) result;
        NodeList childNodes=MethodNodes.item(0).getChildNodes();
        JOptionPane.showMessageDialog(null, "No of children of "+path+" is "+childNodes.getLength());
        for(int i=0;i&lt;=childNodes.getLength()-1;i++)
        {
            Node node=childNodes.item(i);
            JOptionPane.showMessageDialog(null, "Found Child "+node.getNodeName());
            traverse(node,parent);
        }

        } catch (XPathExpressionException e) {
        JOptionPane.showMessageDialog(null,"error: "+e.toString());
        e.printStackTrace();
    }
}
private void traverse(Node root ,CreateFlowChart parent)
{
    if(root.getNodeName()=="If")
    {
        String condition="";
        NodeList childNodes=root.getChildNodes();
        for(int i=0;i&lt;childNodes.getLength();i++)
        {
            Node child=(Node)childNodes.item(i);
            if(child.getNodeName()=="condition")
            {
                Element ele=(Element)child;
                condition=ele.getTextContent();
            }
        }
        dataObjects.ControlStatements ifstmt=new dataObjects.ControlStatements(condition,null,true);
        parent.foundIf(ifstmt);
        NodeList childs=root.getChildNodes();
        for(int i=0;i&lt;childs.getLength();i++)
        {
            Node child=(Node)childs.item(i);
            traverse(child,parent);

        }
        parent.foundEndIf();

    }
    else if(root.getNodeName()=="else")
    {
        parent.foundElse();
        NodeList childNodes=root.getChildNodes();
        for(int i=0;i&lt;childNodes.getLength();i++)
        {
            Node child=(Node)childNodes.item(i);
            traverse(child,parent);
        }
        parent.foundEndElse();
    }
    else if(root.getNodeName()=="Statement")
    {
        parent.foundStatement(root.getTextContent());
    }
}

现在从读取器调用的三个函数是:

public void foundIf(dataObjects.ControlStatements Ifstatement)
{
    JOptionPane.showMessageDialog(null,"Drawing If");
    graph.getModel().beginUpdate();
    try
    {
        Object v1 = graph.insertVertex(start, null, "If "+Ifstatement.condition, 20, 20, 150,60,"Branch");
        if(isInIf==false && isInElse==false)
        {
            JOptionPane.showMessageDialog(null,"Drawing normally");
            graph.insertEdge(start, null, "", currentNode,v1);
        }
        else if(isInIf==true)
        {
            JOptionPane.showMessageDialog(null,"Drawing inside a previous If");
            graph.insertEdge(start, null, "True", currentNode,v1);
            isInIf=false;
        }
        else if(isInElse==true)
        {
            JOptionPane.showMessageDialog(null,"Drawing inside a previous else");
            graph.insertEdge(start, null, "False", currentNode,v1);
            isInElse=false;
        }
        currentNode=(mxCell)v1;
        JOptionPane.showMessageDialog(null,"Pushing if node inside stack");
        lastIfNode.push(currentNode);
        isInIf=true;
    }
    finally
    {
        graph.getModel().endUpdate();
    }
}
public void foundElse()
{
    currentNode=lastIfNode.pop();
    isInElse=true;
}
public void foundStatement(String st)
{
    JOptionPane.showMessageDialog(null,"Drawing a statement");
    graph.getModel().beginUpdate();
    try
    {
        Object v1 = graph.insertVertex(currentNode, null, st, 20, 20, 150,60,"Statement");
        if(isInIf==false && isInElse==false)
        {
            JOptionPane.showMessageDialog(null,"Drawing normally");
            graph.insertEdge(start, null, "", currentNode,v1);
        }
        else if(isInIf==true)
        {
            JOptionPane.showMessageDialog(null,"Drawing inside a prv If");
            graph.insertEdge(start, null, "True", currentNode,v1);
            isInIf=false;
        }
        else if(isInElse==true)
        {
            JOptionPane.showMessageDialog(null, "Drawing inside else");
            graph.insertEdge(start, null, "False", currentNode,v1);
            isInElse=false;
            reTraceIf=false;
        }
        if(reTraceIf==true)
        {
            JOptionPane.showMessageDialog(null, "Drawing false part");
            graph.insertEdge(start, null, "False", lastIfNode.pop(),v1);
            reTraceIf=false;
        }
        if(branchEnded==true)
        {
            JOptionPane.showMessageDialog(null, "Linking brancehs");
            graph.insertEdge(start, null, "", prvBranchNode,v1);
            branchEnded=false;
        }
        currentNode=(mxCell)v1;
    }
    finally
    {
        graph.getModel().endUpdate();
    }
}
public void foundEndIf()
{
    prvBranchNode=currentNode;
    isInIf=false;
    reTraceIf=true;
}
public void foundEndElse()
{
    branchEnded=true;
}

这对于 if else 语句很有效,但之后我会恶化,我知道这是因为 globbal 变量 prvNode 一次只能 hld 一个节点,可能是一个列表 ll 做,但仍然可能会出现一些问题......任何人都可以改进它吗?

你可以做一个递归方法来做一个DFS遍历,例如 -

1    For each line in block
2        If it is an if statement
3            Render condition in a Diamond shape
4            Get the statement block inside the if statement and go to #1
5        else 
6            Render it in a rectangle shape

这只是一个想法,除了记住 2-D 平面上元素的布局外,您还需要跟踪深度。但是您应该先制定遍历算法,然后再拟合布局。

相关内容

  • 没有找到相关文章

最新更新