如何正確釋放分配給鏈表的內存？

Question

我正在寫一個使用鏈接列表的程序，在我的代碼中，我使用「new」關鍵字創建了一個對象指針。據我所知，任何時候你使用新的關鍵字，你也需要刪除，我想知道如果我正確地做到這一點。

#include"gameClass.h" 
#include"card.h" 
#include"list.h" 
#include"node.h" 

using namespace std; 

List::List() 
{ 
    head = NULL; 
} 

List::~List() 
{ 
    delete head; 
} 

void List::add(Card* tmpCard) 
{ 
    if (head == NULL) 
{ 
    Node* tmpNode; 
    tmpNode = new Node; 
    tmpNode->setNext(NULL); 
    tmpNode->setData(tmpCard); 
    head = tmpNode; 
    delete tmpNode; 
} 
else 
{ 
    Node* tmpNode; 
    tmpNode = new Node; 
    tmpNode->setNext(head->getNext()); 
    tmpNode->setData(tmpCard); 
    head = tmpNode; 
    delete tmpNode; 
} 
} 

Card * List::remove() 
{ 
    if (head != NULL) 
{ 
    //Card* tmpCard; 
    Node* tmpNode; 

    tmpNode = new Node; 
    tmpNode->setNext(head->getNext()); 
    tmpNode->setData(head->getData()); 
    head->setNext(tmpNode->getNext()); 
    delete tmpNode; 
} 
    return nullptr; 
} 

Answer 1

你這樣做完全錯了。

• 您將引入一個新的head然後delete新head馬上。太糟糕了。
• 您的remove()函數幾乎沒有任何功能。它創建一個新的Node，爲它設置一些參數，然後檢索應該是相同的數據集，並且delete新的Node。

試試這個：

#include"gameClass.h" 
#include"card.h" 
#include"list.h" 
#include"node.h" 

using namespace std; 

List::List() 
{ 
    head = NULL; 
} 

List::~List() 
{ 
    // delete all node currently have instead of only head 
    while (head != NULL) 
    { 
     Node* tmpNode = head; 
     delete head; 
     head = tmpNode; 
    } 
} 

void List::add(Card* tmpCard) 
{ 
    if (head == NULL) 
    { 
     Node* tmpNode; 
     tmpNode = new Node; 
     tmpNode->setNext(NULL); 
     tmpNode->setData(tmpCard); 
     head = tmpNode; 
     // do not delete the new node here! 
    } 
    else 
    { 
     Node* tmpNode; 
     tmpNode = new Node; 
     tmpNode->setNext(head); // the new head should be linked to current head, not the next node of current head 
     tmpNode->setData(tmpCard); 
     head = tmpNode; 
     // do not delete the new node here! 
    } 
} 

Card * List::remove() 
{ 
    if (head != NULL) 
    { 
     //Card* tmpCard; 
     Node* tmpNode; 

     // no creating new nodes to remove the first node 
     tmpNode = head->getNext(); // remember where the next node is 
     delete head; // remove the head 
     head = tmpNode; // move the head 
    } 
    return nullptr; 
} 

UPDATE：如果你想指針返回卡在節點被移除時，remove()功能將是這樣的：

Card * List::remove() 
{ 
    if (head != NULL) 
    { 
     Card* tmpCard; 
     Node* tmpNode; 

     tmpCard = head->getData(); // remember where the card pointed by head is 
     tmpNode = head->getNext(); // remember where the next node is 
     delete head; // remove the head 
     head = tmpNode; // move the head 
     return tmpCard; // return the card 
    } 
    return nullptr; // there are no cards in this list 
} 

Answer 2

要避免此問題，只需使用std::shared_ptr<Node>即可。只要指針超出範圍，指針就會釋放它指向的內容（如果沒有其他共享指針指向它）。

如果您確實想要練習使用new/delete，您會誤解它的作用：使用new創建新節點，然後以某種方式使用該節點。只有當節點不再需要時，你纔會刪除它。現在發生以下情況：

Node* tmpNode; 
tmpNode = new Node;  // You get a new Node on heap. 
tmpNode->setNext(NULL);  
tmpNode->setData(tmpCard); 
head = tmpNode;   // Head now points to the new Node. All is well. 
delete tmpNode;   // The Node both tmpNode and head point to is deleted! 

現在，head和tmpNode都包含一個指向未分配內存的指針。當你訪問它時，你會得到未定義的行爲。你可能想要的是刪除List的析構函數中的所有節點，以及其他任何地方。