C++遞歸樹與指針和狀態機構建

Question

我有一個簡單的狀態機（在下面輸入）。我的主要問題是我試圖對我的狀態機的函數進行遞歸調用。我在輸入函數時所做的是爲我的樹創建一個新節點，然後推送它。當我進行遞歸調用時，我一遍又一遍地創建一個新節點。這可以工作，但是當向父母添加孩子時，我有點困惑。有人可以幫助看看這個，並幫助我把我的樹節點（我假設的父母）和附加一個孩子嗎？

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; 

    } 

你可能會認爲dex只是一個返回正確狀態或11（錯誤）的列表數組的索引。你也可以假設這個函數在輸入文件中至少被調用一次，並且已經開始解析。感謝您的幫助。在你的代碼

Answer 1

看，你有

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; 

我看不出其他原因，就state總是等於1，你的代碼在案件8會失敗。假設treeNodeClass::AddChildren(TreeNodeClass*)已被正確實施。無需這些代碼，我可以認爲這不是你的問題。有沒有一種方法可以使state在交換機中不是1？