如果初始输入为 -1，为什么函数会退出？

Question

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

/*EVERYTHING WORKS EXCEPT FOR INITIAL -1*/

// Implementing a Node Structure - This represents a node of data
typedef struct _listnode
{
    int item; // Data
    struct _listnode *next; // Linkage
} ListNode;

// Core Functions of a Linked List
void printList(ListNode *head);
ListNode* findNode(ListNode *head, int index);
int insertNode(ListNode **ptrHead, int index, int value);
void removeNode(ListNode **ptrHead, int index);


int main()
{
    // Instantiate a Linked List
    ListNode *head = NULL, *temp;
    /*
    head is a pointer variable that will eventually point to the firstNode.
    temp is a pointer variable that points to the lastNode. This is used in from Linked List functions.
    */
    int num_to_store;
    printf("Enter an Integer to Store (-1 to end):\n");
    scanf("%d", &num_to_store);

    while (num_to_store != -1)
    {
        // Initialize a Linked List
        if (head == NULL)
        {
            head = malloc(sizeof(ListNode));
            temp = head;
        }
        else
        {
            temp->next = malloc(sizeof(ListNode));
            temp = temp->next;
        }
        temp->item = num_to_store;
        printf("Enter an Integer to Store (-1 to end):\n");
        scanf("%d", &num_to_store);
    }
    temp->next = NULL;

    // Menu-Driven Application
    int user_choice, index, value, *p, option_3;
    puts("");
    printf("----------------\n 1: printList()\n 2: findNode()\n 3: insertNode()\n 4: removeNode()\n-1: End\n----------------\n");
    scanf("%d", &user_choice);
    while (user_choice != -1)
    {
        switch(user_choice)
        {
        case 1:
            printList(head);
            break;
        case 2:
            printf("Enter index to search:\n");
            scanf("%d", &index);
            p = findNode(head, index);
            if (p != NULL) printf("Node Item: %d\n", *p);
            break;
        case 3:
            printf("Enter index to insert:\n");
            scanf("%d", &index);
            printf("Enter value to insert:\n");
            scanf("%d", &value);
            option_3 = insertNode(&head, index, value);
            if (option_3 == 0) printf("NODE INSERTED!\n");
            break;
        case 4:
            printf("Enter index to remove:\n");
            scanf("%d", &index);
            removeNode(&head, index);
            break;

        }

        // Prompt User to Make Another Selection
        puts("");
        printf("----------------\n 1: printList()\n 2: findNode()\n 3: insertNode()\n 4: removeNode()\n-1: End\n----------------\n");
        scanf("%d", &user_choice);

    }

    return 0;
}

void printList(ListNode *head)
{
    // Linked List is Empty
    if (head == NULL)
    {
        printf("LINKED LIST IS EMPTY!\n");
    }

    // Linked List is Not Empty
    else
    {
        printf("LINKED LIST: ");
        while (head != NULL)
        {
            printf("%d ", head->item);
            head = head->next;
        }
        puts("");
    }
}

ListNode* findNode(ListNode *head, int index)
{
    // Linked List is Empty or Index is Invalid
    if (head == NULL || index < 0)
    {
        printf("INVALID!\n");
        return NULL;
    }

    // Linked List is not Empty, Check if Index > len(Linked List)
    else
    {
        while (index > 0)
        {
            head = head->next;
            if (head == NULL)
            {
                printf("INVALID!\n");
                return NULL;
            }
            index--;
        }
        printf("INDEX FOUND!\n");
        return head;
    }
}

int insertNode(ListNode **ptrHead, int index, int value)
{
    ListNode *cur, *pre;
    // Linked List is Empty || Insert at Index 0
    if ((*ptrHead) == NULL || index == 0)
    {
        cur = *ptrHead;
        *ptrHead = malloc(sizeof(ListNode));
        (*ptrHead)->item = value;
        (*ptrHead)->next = cur;
        return 0;
    }
    // Insert in the Middle
    else
    {
        pre = findNode(*ptrHead, index-1);
        if (pre != NULL)
        {
            cur = pre->next; // This is temporary
            pre->next = malloc(sizeof(ListNode)); // L connects to nN
            pre->next->item = value;
            pre->next->next = cur; // nN connects to R
        }
        return 0;
    }
    return -1;
}

void removeNode(ListNode **ptrHead, int index)
{
    ListNode *cur, *pre;
    // Linked List is Empty
    if ((*ptrHead == 0) || index < 0)
    {
        printf("INVALID!\n");
    }

    // Remove firstNode
    else if (index == 0)
    {
        cur = findNode(*ptrHead, index);
        *ptrHead = cur->next;
        free(cur); // Unallocate the memory
    }

    // Remove Middle
    else
    {
        pre = findNode(*ptrHead, index-1);
        cur = findNode(*ptrHead, index);
        pre->next = cur->next;
        cur->next = NULL;
    }
}

大家好，我正在用C 编写一个具有4 个核心功能的链表程序。除了当我编译代码和类型 -1 时，函数和其他一切都有效。程序退出。我想知道出了什么问题。我尝试在 code::blocks 上进行调试，但没有任何结果。

我想达到什么目的？

创建一个没有节点的链表。
编码链表的4个核心函数(insert/remove/print/search)。

我尝试了什么？

分散 printf() 语句以进行调试。
在 Whosebug 上检查了类似的问题。
查看了 geeksforgeeks 上的代码。

知道问题的原因吗？对于此类问题，有可能的解决方案吗？

Answer 1

当您键入 -1 时，将跳过整个 while (num_to_store != -1) 循环，并且 temp 将保持未初始化状态，这将使 temp->next = NULL; 调用未定义的行为。

如果初始输入为 -1，为什么函数会退出？

Why is function exiting if the initial input is -1?

c

linked-list

data-structures

singly-linked-list