将缓冲区 (char*) 传递给 C 中的函数

Question

我将缓冲区 (char*) 传递给 C 中的函数。

在函数内部，我为缓冲区分配内存并附加一个字符串（来自虚拟服务器的响应）。当在函数内部打印时，字符串将显示为从服务器发送的。

当尝试从函数外部读取缓冲区时，我没有得到我需要的结果：

A。// 字符* server_message; // 不起作用，错误:: 分段错误（核心已转储）

b. // char* server_message = calloc((BUFFER_SIZE) + 1, sizeof(char)); // 不起作用，没有生成错误，下面的 server_message 的 printf 打印一个空值:: ==> server_message:: || ||

C。char* server_message = calloc((1000 * BUFFER_SIZE) + 1, sizeof(char)); // 分配超出所需内存的工作:: ==> server_message:: || 2016-03-08 12:20:13 你好 ||

我根本不喜欢选项 c，因为我分配的内存超出了我的需要。出于测试目的，BUFFER_SIZE 当前设置为 10。

这是调用部分：：

write_to_server(client_socket, message, server_message);
free(message);
printf("\n\n==> server_message:: || %s ||\n\n", server_message);

if (server_message != NULL)
    free(server_message);

这是 write_to_server 函数代码：

void write_to_server(int file_descriptor, char* message, char* server_message)
{

    char* msg = strcat(message, "\n");

int n_bytes = write(file_descriptor, msg, strlen(msg) + 1);

if (n_bytes <= 0)
{
    perror("write");
    exit(EXIT_FAILURE);
}

if (DEBUG_MODE) 
    printf("\nsuccessfully written %d bytes to server:: %s", n_bytes, msg);

// char* message_back = calloc(BUFFER_SIZE + 1, sizeof(char));
if (server_message == NULL)
{
    server_message = calloc(BUFFER_SIZE + 1, sizeof(char));
}

if (server_message == NULL)
{
    perror("Could not allocate memory for server_message");
    exit(EXIT_FAILURE);
}

char* tmp_message_back = calloc(BUFFER_SIZE + 1, sizeof(char));

if (tmp_message_back == NULL)
{
    if (DEBUG_MODE)
        perror("Could not allocate memory for tmp_message_back");

    exit(EXIT_FAILURE);
}

int n_read = 0; 
int n_total_read = 0;

while ((n_read = read(file_descriptor, tmp_message_back, BUFFER_SIZE)) > 0)
{
    n_total_read += n_read;
    tmp_message_back[n_read] = '\0';

    /*
    char* strcat(char* destination, const char* source);

    Appends a copy of the source string to the destination string.
    The terminating null character in destination is overwritten by the first character of source, 
    and a null-character is included at the end of the new string formed by the concatenation of both in destination.
    */
    server_message = strcat(server_message, tmp_message_back);

    // void* realloc(void* ptr, size_t size);
    /*
    char* new_message_back = realloc(server_message, n_total_read + (BUFFER_SIZE * sizeof(char)));

    if (new_message_back == NULL)
    {
        perror("Could not allocate memory for server_message while receiving bytes from server");

        free(tmp_message_back);

        return;
    }

    server_message = new_message_back;
    */
    server_message = realloc(server_message, n_total_read + (BUFFER_SIZE * sizeof(char)));
    server_message[n_total_read] = '\0';

    if (DEBUG_MODE)
        printf("\nread %d (%d in total), size:: %zu ===> | %s |\t| %s |\n", n_read, n_total_read, strlen(server_message), server_message, tmp_message_back);
}

server_message = realloc(server_message, n_total_read);

if (DEBUG_MODE)
{
    printf("\n\n\n\n-- read %d in total, size:: %zu\n", n_total_read, strlen(server_message));
    printf("\n\nserver_message:: %s\n\n", server_message);
}

free(tmp_message_back);
// free(message_back);
}

Answer 1

在 C 语言中，参数是按值传递的。基本上你正在这样做，期望输出是
Hello world!：

void Test(char *ptr)
{
    ptr = malloc(100);   // this modifies ptr but not
                         // somepointer in the main function
    strcpy(ptr, "Hello world!");
}

int main()
{
    char *somepointer;  // somepointer is not initialized and contains
                        // an indetermined value
    Test(somepointer);
                        // here the value of somepointer has not changed

    printf("%s", somepointer);   // here you try to printf a char pointer
                                 // that points to an idetermined location
    return 0;
}

这是未定义的行为，很可能会打印垃圾或崩溃。

你需要的是这样的：

void Test(char **ptr)
{
    *ptr = malloc(100);
    strcpy(*ptr, "Hello world!");
}

int main()
{
    char *somepointer;
    Test(&somepointer);   // pass the pointer to sompointer to Test
    printf("%s", somepointer);  // now somepointer points to the memory
                                // mallocd in Test
    return 0;
}