为什么链表插入节点中“新建”后没有“删除”

Ste*_*eak 1 c++ linked-list new-operator delete-operator

我一直试图通过阅读一些文本和查找内容来理解 C++ 中的内存分配。我经常看到人们应该总是在“new”之后调用“delete”。但是,我也看到这样的代码:

void LinkedList::add(int data){
    Node* node = new Node();
    node->data = data;
    node->next = this->head;
    this->head = node;
    this->length++;
}
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在链表或堆栈等结构中。

我在 SO 上看到了一些很好的解释,例如:

为什么 C++ 程序员应该尽量减少“new”的使用? 在 C++ 中,何时使用“new”,何时不使用?

但是,我仍然很困惑为什么不在这里为新节点调用“删除”。

编辑:让我澄清我的问题。我明白为什么不立即以相同的方法调用 delete 。但是,在相同的代码中,我没有看到与添加匹配的删除语句。我假设一旦程序结束,一切都会被删除,但我很困惑,没有明显匹配的删除语句(即:计算代码中的所有新闻,计算代码中的所有删除,它们不匹配)。

编辑:这是我正在查看的来源:https : //www.geeksforgeeks.org/linked-list-set-2-inserting-a-node/

他们的链表的代码:

// A complete working C++ program to demonstrate
//  all insertion methods on Linked List
#include <bits/stdc++.h>
using namespace std;
 
// A linked list node
class Node
{
    public:
    int data;
    Node *next;
};
 
/* Given a reference (pointer to pointer)
to the head of a list and an int, inserts
a new node on the front of the list. */
void push(Node** head_ref, int new_data)
{
    /* 1. allocate node */
    Node* new_node = new Node();
 
    /* 2. put in the data */
    new_node->data = new_data;
 
    /* 3. Make next of new node as head */
    new_node->next = (*head_ref);
 
    /* 4. move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Given a node prev_node, insert a new node after the given
prev_node */
void insertAfter(Node* prev_node, int new_data)
{
    /*1. check if the given prev_node is NULL */
    if (prev_node == NULL)
    {
        cout<<"the given previous node cannot be NULL";
        return;
    }
 
    /* 2. allocate new node */
    Node* new_node = new Node();
 
    /* 3. put in the data */
    new_node->data = new_data;
 
    /* 4. Make next of new node as next of prev_node */
    new_node->next = prev_node->next;
 
    /* 5. move the next of prev_node as new_node */
    prev_node->next = new_node;
}
 
/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
void append(Node** head_ref, int new_data)
{
    /* 1. allocate node */
    Node* new_node = new Node();
 
    Node *last = *head_ref; /* used in step 5*/
 
    /* 2. put in the data */
    new_node->data = new_data;
 
    /* 3. This new node is going to be
    the last node, so make next of
    it as NULL*/
    new_node->next = NULL;
 
    /* 4. If the Linked List is empty,
    then make the new node as head */
    if (*head_ref == NULL)
    {
        *head_ref = new_node;
        return;
    }
 
    /* 5. Else traverse till the last node */
    while (last->next != NULL)
        last = last->next;
 
    /* 6. Change the next of last node */
    last->next = new_node;
    return;
}
 
// This function prints contents of
// linked list starting from head
void printList(Node *node)
{
    while (node != NULL)
    {
        cout<<" "<<node->data;
        node = node->next;
    }
}
 
/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
     
    // Insert 6. So linked list becomes 6->NULL
    append(&head, 6);
     
    // Insert 7 at the beginning.
    // So linked list becomes 7->6->NULL
    push(&head, 7);
     
    // Insert 1 at the beginning.
    // So linked list becomes 1->7->6->NULL
    push(&head, 1);
     
    // Insert 4 at the end. So
    // linked list becomes 1->7->6->4->NULL
    append(&head, 4);
     
    // Insert 8, after 7. So linked
    // list becomes 1->7->8->6->4->NULL
    insertAfter(head->next, 8);
     
    cout<<"Created Linked list is: ";
    printList(head);
     
    return 0;
}
 
 
// This code is contributed by rathbhupendra
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Nic*_*las 8

您引用的代码应该delete是某个节点的节点。事实上,该代码展示了大量糟糕的 C++ 实践。它不会删除节点,因为它是错误的代码。

哦,顺便说一句:忽略您链接到的网站上的任何内容。如果该站点上有一些有用的东西,那只是偶然。

  • 很高兴知道。我可能会咬紧牙关,而是完全阅读一些书。 (2认同)