是否有適當的方法來檢查單鏈表的尾節點是否已被分配了一個有意義的值？

通常，在我們的單鏈表推送函式中，我們必須檢查我們是否正在查看尾節點，以及它是否已被賦值，對嗎？因為如果它是尾節點還沒有被賦值，我們不應該創建一個新節點；我們必須簡單地為其存盤變數分配一個值。在其余情況下 - 我們必須在分配之前創建一個。

但是檢查尾部（具有垃圾值/零/未使用/用戶未觸及）的正確方法是什么？ 或者是否有其他所有人都在使用的其他方法來解決所有這些問題（包括在推入 SLL 時將未使用的尾部視為特殊情況） ？

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>

struct ListNode
{
    int val;
    struct ListNode *next;
};

void linked_init(struct ListNode **p)
{
    (*(p)) = malloc(sizeof(struct ListNode));
    (*(p))->next = NULL;
    //Basically saying that for every linked list that I'll create and initialize, using the "ListNode" struct definition, I will assign to...
    //...its Tail Node, a "Unique Code (int value)", which will indicate that no SENSICAL value had been assigned to the tail, yet.
    (*(p))->val = 98989;
}

void linked_push(struct ListNode **p, int curval)
{
    //If it's the tail, and within the tail, the storage hasn't been used - aka - a usable int value hasn't been assigned.
    if ((*(p))->next == NULL && (*(p))->val == 98989)
    {
        //Push it into the already existing Tail Node
        (*(p))->val = curval;
    }
    else
    {
        //It's either not a tail, or it is indeed a tail which has already been assigned a value.
        //Therefore, we must create a new node before assigning into its tail node, a value.
        struct ListNode *temp;
        temp = malloc(sizeof(struct ListNode));
        temp->val = curval;
        temp->next = (*(p));
        (*(p)) = temp;
    }
}

void linked_display(struct ListNode *p)
{
    struct ListNode *temp;
    temp = p;
    while (temp)
    {
        printf("%d", (temp)->val);
        printf("\n");
        temp = temp->next;
    }
}

int main(int argc, char **argv)
{
    struct ListNode *list;
    linked_init(&list);

    // Initial push would be targeted at the tail, since we were the ones who initialized it; using the linked_init() function.
    // This whole thing only works if the linked list belonged to us; if we initialized it.
    // If we are applying these functions on a list imported from external sources - ex. from questions on online coding platforms, etc - then this wouldn't work.
    linked_push(&list, 14);
    linked_push(&list, 17);
    linked_push(&list, 20);
    linked_push(&list, 90);

    linked_display(list);

    return 0;
}

這就是我目前的做法，但我認為這不是一種實用的方法。我在網上找不到有關此問題的任何資訊。

uj5u.com熱心網友回復：

沒有必要分配一個具有魔法值的節點作為尾節點。空串列只是一個空指標。使用哨兵值是令人困惑和容易出錯的：應該忽略提倡這種方法的教程。

這是一個簡化版本：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

void linked_init(struct ListNode **p) {
    *p = NULL;
}

void linked_push(struct ListNode **p, int curval) {
    // allocate a new node and insert it at the head of the list
    struct ListNode *temp = malloc(sizeof(*temp));
    if (temp) {
        temp->val = curval;
        temp->next = *p;
        *p = temp;
    }
}

int linked_pop(struct ListNode **p) {
    struct ListNode *temp = *p;
    int val = 0;
    if (temp) {
        val = temp->val;
        *p = temp->next;
        free(temp);
    }
    return val;
}

void linked_display(const struct ListNode *p) {
    const struct ListNode *temp;
    for (temp = p; temp; temp = temp->next) {
        printf("%d ", temp->val);
    }
    printf("\n");
}

int main(int argc, char **argv) {
    struct ListNode *list;

    // Initialize the list. Could just write: struct ListNode *list = NULL;
    linked_init(&list);

    // Insert values
    linked_push(&list, 14);
    linked_push(&list, 17);
    linked_push(&list, 20);
    linked_push(&list, 90);

    // Print the list values: 90 20 17 14
    linked_display(list);

    // Free the list.
    while (list) {
        linked_pop(&list);
    }

    return 0;
}

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