鏈表的C 指標演算法

我剛剛開始自學 C ，作為一個玩具問題，我正在嘗試執行以下操作 - 給定一個鏈表，我想將所有節點存盤到一個新串列中，并回傳這個新串列。對于背景關系，我來自 Python 背景。

我有以下程式 -

#include <iostream>
using namespace std;

struct node
{
    unsigned val;
    struct node *next;
};

node *even_nodes(node **root)
{
    node *new_list_head = NULL;
    node *new_list_runner = NULL;
    node *runner = *root;
    while (runner != NULL)
    {
        if (new_list_head != NULL){
            printf("OUTSIDE LOOP new_list_head.val = %d\n", new_list_head->val);
        }
        if (runner->val % 2 == 0)
        {
            cout << runner->val << endl;
            node new_node = {.val = runner->val, .next = NULL};
            if (new_list_head == NULL)
            {
                printf("new_list_head is NULL!\n");
                new_list_head = &new_node;
                new_list_runner = &new_node;
                printf("after allocation. new_list_head.val = %d\n", new_list_head->val);
            }
            else
            {
                printf("new_list_head is NOT NULL! new_list_head.val = %d\n", new_list_head->val);
                new_list_runner->next = &new_node;
                new_list_runner = new_list_runner->next;
                printf("after allocation. new_list_head.val = %d\n", new_list_head->val);
            }
        }
        runner = runner->next;
    }
    printf("new_list_head val = %d\n", new_list_head->val);
    return new_list_head;
};

void add_data(node **root, int new_data)
{
    node *new_node = (node *)malloc(sizeof(node *));
    new_node->val = new_data;
    new_node->next = (*root);
    (*root) = new_node;
}

void print_list(node *root)
{
    node *head = root;
    while (head != NULL)
    {
        printf("%d -> ", head->val);
        head = head->next;
    }
    printf("END\n");
};

int main()
{
    node *head = NULL;
    add_data(&head, 19);
    add_data(&head, 18);
    add_data(&head, 3);
    add_data(&head, 4);
    add_data(&head, 1);
    printf("Initial list:\n");
    print_list(head);
    node *new_list = even_nodes(&head);
    printf("New list of even numbers: \n");
    print_list(new_list);

    return 0;
}

輸出如下 -

Initial list:
1 -> 4 -> 3 -> 18 -> 19 -> END
4
new_list_head is NULL!
after allocation. new_list_head.val = 4
OUTSIDE LOOP new_list_head.val = 4
OUTSIDE LOOP new_list_head.val = 4
18
new_list_head is NOT NULL! new_list_head.val = 18
after allocation. new_list_head.val = 18
OUTSIDE LOOP new_list_head.val = 18
new_list_head val = 18
New list of even numbers: 
[1]     segmentation fault 

我不明白為什么new_list_head也隨著改變new_list_runner？為什么我new_list_head指向新串列的最后一個元素，而不是第一個？

另外，為什么會出現段錯誤？方法中print_list，為什么是守衛

while (head != NULL)

不作業？

任何幫助將不勝感激！

uj5u.com熱心網友回復：

您不能通過獲取區域變數的地址來進行動態分配。當作用域退出時，區域變數被銷毀，你留下一個指向不存在的物件的指標（稱為懸空指標）。

你的代碼在這里有這個問題

node new_node = {.val = runner->val, .next = NULL}; // local variable
if (new_list_head == NULL)
{
    new_list_head = &new_node; // BAD
    new_list_runner = &new_node; // BAD
}
else
{
    new_list_runner->next = &new_node; // BAD
    new_list_runner = new_list_runner->next;
}

相反，您應該使用它new來分配新節點，使用它們創建的物件在new您使用它們之前不會被銷毀delete。

node* new_node = new node{runner->val, NULL};
if (new_list_head == NULL)
{
    new_list_head = new_node;
    new_list_runner = new_node;
}
else
{
    new_list_runner->next = new_node;
    new_list_runner = new_list_runner->next;
}

標籤：C 指针链表

上一篇：C 中指標的靜態優先級佇列

下一篇：求2個單鏈表的交集和并集