c - 单链表

Question

我创建了一个单链表。一切正常。

我只想知道我是否在我的代码中做了任何有潜在危险的事情。我关心的代码片段是我的推送、弹出和清理。部分代码仅用于用户交互，因此并不重要(无论如何我都发布了它，以便更清楚地了解我在做什么)。只是链表应用。

非常感谢任何建议，因为这是我的第一次尝试。

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

typedef struct product_data product_data_t;
struct product_data
{
    int product_code;
    char product_name[128];
    int product_cost;
    product_data_t *next;
};

static product_data_t *head = NULL;
static product_data_t *tail = NULL;
static product_data_t *new_product = NULL;

// Push a product on to the list.
void push(int code, char name[], int cost); 
// Pop (delete) a product from the list.
void pop(int code);
// Display all product in the list.
void display_list();
// Delete all memory allocated on the list
void clean_up();
// Display menu
void menu();

int main(void)
{
    menu();

    getchar();

    return 0;
}

void push(int code, char name[], int cost)
{
    // Allocate memory for the new product
    new_product = calloc(1, sizeof(product_data_t));
    if(!new_product)
    {
        fprintf(stderr, "Cannot allocated memory");
        exit(1);
    }

    /* Populate new products elements fields */
    new_product->product_code = code;
    strncpy(new_product->product_name, name, sizeof(new_product->product_name));
    new_product->product_cost = cost;
    new_product->next = NULL;

    // Set the head and tail of the linked list
    if(head == NULL)
    {
        // First and only product
        head = new_product;
    }
    else
    {
        tail->next = new_product;
    }

    tail = new_product;
}

// Find the product by code and delete
void pop(int code)
{
    product_data_t *product = head;
    product_data_t *temp = NULL;
    product_data_t *previous = head;
    int found = 0; // 0 - Not Found, 1 - Found

    if(!head)
    {
        puts("The list is empty");
        return;
    }

    while(product)
    {
        if(product->product_code == code)
        {
            found = 1; // Found
            // Check if this is in the first node - deleting from head
            if(head->product_code == code)
            {
                temp = head;
                head = head->next;
                free(temp);

                // Finished Deleting product
                return;
            }

            // Check if this is the end node - deleting from the tail
            if(tail->product_code == code)
            {
                temp = tail;
                tail = previous;
                free(temp);

                // Finished deleting product
                return;
            }

            // delete from list if not a head or tail
            temp = product;
            previous->next = product->next;
            free(temp);

            // Finished deleting product
            return;
        }
        // Get the address of the previous pointer.
        previous = product;
        product = product->next;  
    }

    if(!found)
    {
        printf("code [ %d ] was not found\n", code);
    }

    // Set all to null after finished with them
    product = NULL;
    temp = NULL;
    previous = NULL;
}

// Traverse the linked list
void display_list()
{
    // Start at the beginning
    product_data_t *product = head;

    while(product)
    {
        printf("===================================\n");
        printf("Product code: \t\t%d\n", product->product_code);
        printf("Product name: \t\t%s\n", product->product_name);
        printf("product cost (USD): \t%d\n", product->product_cost);
        printf("===================================\n\n");

        // Point to the next product
        product = product->next;
    }
    // Finished set to null
    product = NULL;
}

// Release all resources
void clean_up()
{    
    product_data_t *temp = NULL;

    while(head)
    {
        temp = head;
        head = head->next;
        free(temp);    
    }
    head = NULL;
    temp = NULL;

    // End program - goodbye
    exit(0);
}

void menu()
{
    int choice = 0, code = 0, cost = 0;
    char name[128] = {0};

    do
    {
        fflush(stdin); // Flush the input buffer

        puts("========= Welecome to linked list ===============");
        puts("[1] Add new product to the list");
        puts("[2] Delete a product from the list");
        puts("[3] Display all products");
        puts("[4] Exit and clean up");
        printf("Enter your choice: ");
        scanf("%d", &choice);

        switch(choice)
        {
        case 1:
            printf("Enter product code: ");
            scanf("%d", &code);
            printf("Enter cost: ");
            scanf("%d", &cost);
            printf("Enter name: ");
            scanf("%s", name);
            push(code, name, cost);
            break;

        case 2:
            printf("Enter product code: ");
            scanf("%d", &code);
            pop(code);
            break;

        case 3:
            display_list();
            break;

        case 4:
            clean_up();
            break;

        default:
            puts("Incorrect choice");
            break;
        }
    }while(choice != 4);
}

Answer 1

来自 pop()

            if(head->product_code == code)
            {
                    temp = head;
                    head = head->next;
                    free(temp);

                    // Finished Deleting product
                    return;
            }

在只有一项的情况下，'head' 和 'tail' 将指向同一个节点。但是，如果弹出这一项，“head”将被调整，但“tail”仍将指向空闲的节点。这将留下一个错误的指针，这可能会导致您的计算机爆炸。

附录:类似地，如果您弹出被推送的最后一个节点，“new_product”将悬空，而 clean_up() 也会使“tail”指针悬空。即使提供的代码示例在它们被释放后永远不会取消引用它们，C 代码中的悬空指针也应始终被视为“潜在危险”。