試圖刪除頭時有意刪除整個鏈表

Question

我正在爲我的C++類開發一個檢查器的模擬遊戲。我的問題在於保存跳棋的鏈接列表。除了列表頭之外，我可以完全刪除任何檢查器。我查看過這裏和其他網站，我相信有某處存在內存泄漏。我對C++相當陌生，所以除了玩弄事物（這可能只會造成更大的問題）之外，我不確定要做些什麼。我從來沒有在這裏發佈過，所以如果格式稍微偏離或太雜亂，請原諒我。我會盡量簡短。首先，這是鏈表的節點類的一個片段。

class CheckerpieceNode 
{ 
private: 
    Checkerpiece *Node; 
    CheckerpieceNode *Next; 
public: 
    CheckerpieceNode(); // sets Node and Next to NULL in .cpp file 
    void setNode(Checkerpiece *node); 
    void setNext(CheckerpieceNode *next); 
    Checkerpiece* getNode(); 
    CheckerpieceNode* getNext(); 
}; 

這些函數的設置與您在Checkerpiece.cpp類中所期望的幾乎相同。 以下是代碼的使用方法。它由我的主類中的Checkerboard對象調用。

theCheckerboard.removeChecker(theCheckerboard.findChecker(selector->getCurrentX() + 0, selector->getCurrentY() - VERTICAL_SHIFT, listHead), listHead); 

VERTICAL_SHIFT只是我的棋盤圖形在控制檯上的方式。由於它適用於所有其他節點（不包括頭部），所以我排除它作爲錯誤的來源。選擇器是棋子對象，但它不是列表的一部分。

以下是Checkerboard類中的實際findChecker和removeChecker代碼。

Checkerpiece* findChecker(int x, int y, CheckerpieceNode* list_head) 
{ 
if(list_head== NULL) return NULL; // do nothing 
else 
{ 
    CheckerpieceNode* node = new CheckerpieceNode; 
    node = list_head; 
    while(node != NULL && node->getNode() != NULL) 
    { 
     if()// comparison check here, but removed for space 
     { 
      return node->getNode(); 
      delete node; 
      node = NULL; 
     } 
     else // traversing 
      node = node->getNext(); 
    } 

    return NULL; 
} 
} 


void removeChecker(Checkerpiece* d_checker, CheckerpieceNode* list_head) 
{ 
if(list_head== NULL) // throw exception 
else 
{ 
    CheckerpieceNode *temp = NULL, *previous = NULL; 
    Checkerpiece* c_checker= new Checkerpiece; 
    temp = list_head; 
    while(temp != NULL && temp->getNode() != NULL) 
    { 
     c_checker= temp->getNode(); 
     if(d_checker!= c_checker) 
     { 
      previous = temp; 
      temp = temp->getNext(); 
     } 
     else 
     { 
      if(temp != list_head) 
      { 
       previous->setNext(temp->getNext()); 
       delete temp; 
       temp = NULL; 
      } 
      else if(temp == list_head) // this is where head should get deleted 
      { 
       temp = list_head; 
       list_head= list_head->getNext(); 
       delete temp; 
       temp = NULL; 
      } 


      return; 
     } 
    } 
} 
} 

Answer 1

噢，我的，你是複雜的。大量冗餘檢查，分配和不必要的變量（如c_checker也會泄漏內存）。

// Write down the various scenarios you can expect first: 
// (a) null inputs 
// (b) can't find d_checker 
// (c) d_checker is in head 
// (d) d_checker is elsewhere in the list 
void removeChecker(Checkerpiece* d_checker, CheckerpieceNode* list_head) { 
    // first sanitize your inputs 
    if (d_checker == nullptr || list_head == nullptr) // use nullptr instead of NULL. its a keyword literal of type nullptr_t 
     throw exception; 

    // You understand that there is a special case for deleting head. Good. 
    // Just take care of it once and for all so that you don't check every time in the loop. 
    CheckerpieceNode *curr = list_head; 

    // take care of deleting head before traversal 
    if (d_checker == curr->getNode()) { 
     list_head = list_head->next; // update list head 
     delete curr; // delete previous head 
     return; // we're done 
    } 

    CheckerpieceNode *prev = curr; 
    curr = curr->next; 

    // traverse through the list - keep track of previous 
    while (curr != nullptr) { 
     if (d_checker == curr->getNode()) { 
      prev->next = curr->next; 
      delete curr; 
      break; // we're done! 
     } 

     prev = curr; 
     curr = curr->next; 
    } 
} 

我希望有幫助。花時間將問題分解成更小的部分，找出可能的情況，如何處理它們，然後纔開始編寫代碼。

Answer 2

removeChecker不能修改list_head的值，因爲它是按值過去的。該方法的簽名應該是：

void removeChecker(Checkerpiece* d_checker, CheckerpieceNode** list_head) 
// You will need to call this function with &list_head 

或

void removeChecker(Checkerpiece* d_checker, CheckerpieceNode* &list_head) 
// Calling code does not need to change 

Answer 3

基於this edit對這個問題筆者，他所使用的解決方案是：

我修改了代碼，以顯示傳入檢查地址刪除 功能。

void delete_checker(Checker* d_checker, CheckerNode* &list_head) // pass by address 
{ 
if(list_head== NULL) // throw exception 
else 
{ 
    CheckerNode*temp = NULL, *previous = NULL; 
    Checker* c_checker= new Checker; 
    temp = list_head; 
    while(temp != NULL && temp->node!= NULL) 
    { 
     c_checker= temp->node; 
     if(d_checker!= c_checker) 
     { 
      previous = temp; 
      temp = temp->next; 
     } 
     else 
     { 
      if(temp != list_head) 
      { 
       previous->next = temp->next; 
       delete temp; 
       temp = NULL; 
      } 
      else if(temp == list_head) // this is where head should get deleted 
      {  
       temp = list_head; 
       list_head= list_head->next; 
       delete temp; 
       temp = NULL; 
      } 

        delete c_checker; 
        c_checker = nullptr; 

      return; 
     } 
    } 
} 
}