C: 如何将经过 malloc 处理的指针传递给函数并释放它们？

Question

我正在开发一个使用套接字和线程的客户端-服务器程序。我有一个客户端菜单，我试图从客户端向服务器发送一个字符，然后调用相应的函数。我正在尝试使用指针和 malloc，但我认为我不太了解如何传递和释放它们。我遇到内存泄漏和错误，例如:

./server': double free or corruption (fasttop): 0x00007f37dc000a50 ***

和

Segmentation fault

这是我的客户端菜单:

char *choice = (char*) malloc(sizeof(char));

do {

    // get a string from the server
    get_msg(sockfd);

    printf("\nFile and Information System\n");
    printf("===========================\n");
    printf("1. Show IP and Student ID\n");
    printf("2. Display server time\n");
    printf("3. Display system information\n");
    printf("4. List files on server\n");
    printf("5. File Transfer\n");
    printf("6. Exit\n");
    printf("Enter choice: ");
    scanf("%s", choice);

    //Send input to server
    send_menu_choice(sockfd, choice);

    switch (*choice) 
    {
        case '1':
                get_ip(sockfd);
            break;
        case '2':
                get_time(sockfd);
            break;
        case '3':
                get_and_send_uname(sockfd);
            break;
        case '4':
                get_file_list(sockfd);
            break;
        case '5':
                printf("File transfer not yet implemented\n");;
            break;
        case '6':
                printf("Exiting...\n");
            break;
        default:
                printf("Invalid choice\n");
    }

    //if(menu_choice != NULL)
    //{
    //    free(menu_choice);
    //}

} while (*choice != '6');

if (choice != NULL)
{
    free(choice);
}

将菜单选项发送到服务器:

void send_menu_choice(int socketNumber, char *choice)
{
    printf("Sending menu choice...\n");
    size_t n = strlen(choice) + 1;

    writen(socketNumber, (unsigned char *) &n, sizeof(size_t));
    writen(socketNumber, (unsigned char *) choice, n);

    printf("Sent choice: %s\n\n", choice);
}

服务器端:

char *menu_choice = (char*) malloc(sizeof(char));

do { 

    printf("Waiting for client to select option...\n\n");
    send_msg(connfd);

    get_menu_choice(connfd, menu_choice);
    printf("Client %d choice was: %s\n", connfd, menu_choice);

    switch (*menu_choice) 
    {
        case '1':
                send_ip(connfd);
            break;
        case '2':
                send_time(connfd);
            break;
        case '3':
                get_and_send_uname(connfd, *uts);
            break;
        case '4':
                send_file_list(connfd);
            break;
        case '5':
                printf("File Transfer not implemented\n");
            break;
        case '6': 
            break;
        default:
                printf("Invalid choice\n");
    }

    //if(menu_choice != NULL)
    //{
    //    free(menu_choice);
    //}

} while (*menu_choice != '6');

if (choice != NULL)
{
    free(choice);
}

从客户端获取菜单选项:

void get_menu_choice(int socketNumber, char *choice)
{
    size_t n;

    readn(socketNumber, (unsigned char *) &n, sizeof(size_t));  
    readn(socketNumber, (unsigned char *) choice, n);

    printf("Received: %zu bytes\n\n", n);
}

---

要求的附加功能:

发送消息(服务器端):

void send_msg(int socket)
{
    char msg_string[] = "\nPlease enter an option:";

    size_t n = strlen(msg_string) + 1;
    writen(socket, (unsigned char *) &n, sizeof(size_t));   
    writen(socket, (unsigned char *) msg_string, n);
}

获取消息(客户端:

void get_msg(int socket)
{
    char msg_string[32];
    size_t k;

    readn(socket, (unsigned char *) &k, sizeof(size_t));    
    readn(socket, (unsigned char *) msg_string, k);

    printf("%s\n", msg_string);
}

我在这里做错了什么？我不会影响指针指向的值吗？

Answer 1

在 do 循环中，您反复释放指针

if(menu_choice != NULL)
{
    free(menu_choice);
}

但是你只在循环之前分配一次。