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我有一个简单的状态机（在下面输入）。我的主要问题是我试图对作为我的状态机的函数进行递归调用。我在输入该函数时所做的是为我的树创建一个新节点，然后将其推送。当我进行递归调用时，我会一遍又一遍地创建一个新节点。这可以工作，但是在将孩子添加到父母时，我有点困惑。有人可以帮忙看看这个并帮我拿我的树节点（我假设是父节点）并将一个孩子附加到它上面吗？

TreeNodeClass* ProcessTree(TokenT token, vector <list <stateAssoc> >& vecTree, int depth)
  {
    int state = 1; //Assume this is a new record.
    bool noState = false;
    bool update = true;
    int dex = 0;
    string root, value, data, tag;
    TreeNodeClass* treeNode;

    treeNode = new TreeNodeClass; //Assume a new node per state machine visit.

    //Need 11 to break out of loop as well. 
    while(state != 10)
    {
      switch(state)
      {
    case 1: dex = 1;
        break;

    case 2: dex = 6; 
        root = yylval;
        break;

    case 3: dex = 7; 
        break;

    case 4: dex = 3;
        value = yylval;
        treeNode->CreateAttrib(root, value);
        break;

    case 5: dex = 2;
        break;

    case 6: dex = 4;
            data = yylval; 
        break;

    case 7: //Really Don't do anything. Set the tag creation at 8...
            dex = 8; 
        tag = yylval;
        if(data != "" and data != "authors")
          treeNode->CreateTag(data, tag);
        break;

    case 8: {
          //New TreeNode already grabbed. 
          //TreeNodeClass* childNode = new TreeNodeClass;
          childNode = ProcessTree(token, vecTree, depth+1);
          childNode->SetHeight(depth);
          treeNode->AddChildren(childNode);
        }
        token = TokenT(yylex()); //Get a new token to process.
        dex = 5;
        break;

    case 9: dex = 9;
        update = false;
        if(yylval != treeNode->ReturnTag())
        {
          state = 11; 
        }
        break;

    case 10: update = false;
        treeNode->SetHeight(1);
        break;

    default: cout << "Error " << endl;
        cout << state << endl;
        cin.get();
        break;

      }

      if(!noState)
    state = FindMatch(vecTree[dex], token);

      if(update)
    token = TokenT(yylex());
      else
    update = true;
    }
    return treeNode;

  }

您可以假设 dex 只是一个列表数组的索引，它将返回正确的状态或 11（错误）。你也可以假设这个函数至少在输入文件上被调用过一次并且已经开始解析。谢谢您的帮助。

4

1 回答 1

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查看您的代码，您有

int state = 1;
//Code

while(state != 10) {
    switch(state) {
        case:1 dex = 1; break;   //Only case you run
        //more cases
        case 8: { //never enter here
          //New TreeNode already grabbed. 
          //TreeNodeClass* childNode = new TreeNodeClass;
          childNode = ProcessTree(token, vecTree, depth+1);
          childNode->SetHeight(depth);
          treeNode->AddChildren(childNode); //should work if assuming function is correct
        }
        //stuff
        break;
        default: break;//blah
    }
}
//blah

return treeNode;

state除了事实总是等于 1之外，我没有看到其他原因，即您在案例 8 中的代码将失败。这是假设treeNodeClass::AddChildren(TreeNodeClass*)已正确实施。如果没有该代码，我可以假设这不是您的问题。是否有某种方法在state开关中不会是 1？

于 2013-08-30T12:48:35.977 回答