C內存映射結構數組泄漏

Question

我有兩個文件，我試圖通過共享內存IPC。我在兩個文件中使用類似的語句進行分配。在服務器文件：

int fd; 
int size = MAX_LEN; 
int bigSize = sizeof(struct region)+ size * sizeof(struct client_message) + size * sizeof(struct server_message); 
    struct region *rptr = (struct region*)malloc(bigSize); 
    printf("region size: %d clientMessage size: %d serverMessage size: %d and the total size: %d\n", sizeof(struct region), size * sizeof(struct client_message), size * sizeof(struct server_message), bigSize); 
    fd = shm_open("/myregion", O_CREAT | O_RDWR, S_IRUSR | S_IWUSR); 

    if (ftruncate(fd, bigSize) == -1) 
     printf("error creating ftruncate\n"); 

    rptr = mmap(NULL, sizeof(struct region), 
       PROT_READ | PROT_WRITE, MAP_SHARED, fd, 0); 

我用下面的鏈接中的示例動態地分配共享存儲器： C Windows - Memory Mapped File - dynamic array within a shared struct

當我在此服務器while循環和增加Ĵ分配

struct client_message { 
    pthread_t client_id; 
    int question; 
}; 
struct server_message { 
    pthread_t client_id; 
    pid_t server_id; 
    int answer; 
}; 
struct region {  /* Defines "structure" of shared memory */ 
    int len; 
    struct client_message ptr_client_message[0]; 
    struct server_message ptr_server_message[0]; 
}; 

文件，

(rptr->ptr_client_message[(j)%size]).question = 30; 
(rptr->ptr_server_message[(j)%size]).answer = 20; 

我從客戶端文件中讀取它是這樣：

printf("rptr len is %d and question of client %d is: %d, answer of server is %d \n", size, k%size, (rptr->ptr_client_message[(k)%size]).question, (rptr->ptr_server_message[(k)%size]).answer); 

輸出被mindboggling： 從服務器終端獲得：

rptr len is 10 and question of client 0 is: 30, answer of server is 20 
rptr len is 10 and question of client 1 is: 30, answer of server is 20 
rptr len is 10 and question of client 2 is: 30, answer of server is 20 
... 

改變用於client_message陣列的10個元素，即，高達客戶端[MAX_LEN]

從客戶終端我得到：

rtpr len is 10 and question of client 0 is: 30, answer of server is 20 
rptr len is 10 and question of client 1 is: 30, answer of server is 30 
rptr len is 10 and question of client 2 is: 30, answer of server is 20 
rptr len is 10 and question of client 3 is: 30, answer of server is 30 
rptr len is 10 and question of client 4 is: 30, answer of server is 20 
rptr len is 10 and question of client 5 is: 30, answer of server is 30 
rptr len is 10 and question of client 6 is: 30, answer of server is 20 
rptr len is 10 and question of client 7 is: 30, answer of server is 20 
rptr len is 10 and question of client 8 is: 30, answer of server is 20 
rptr len is 10 and question of client 9 is: 30, answer of server is 20 
rptr len is 10 and question of client 0 is: 30, answer of server is 20 
rptr len is 10 and question of client 1 is: 30, answer of server is 30 
rptr len is 10 and question of client 2 is: 30, answer of server is 20 
rptr len is 10 and question of client 3 is: 30, answer of server is 30 
rptr len is 10 and question of client 4 is: 30, answer of server is 20 
rptr len is 10 and question of client 5 is: 30, answer of server is 30 
rptr len is 10 and question of client 6 is: 30, answer of server is 20 
rptr len is 10 and question of client 7 is: 30, answer of server is 20 
rptr len is 10 and question of client 8 is: 30, answer of server is 20 

因此，從另一個進程到達時，struct區域中的條目是混合的。我怎樣才能防止這一點？

Answer 1

問題是你不能在一個結構中使用兩次zero length array trick。

您在struct region中使用的技巧是GCC擴展，它允許您在結構中包含可變長度的數組。零長度數組作爲標題，然後你可以記住，你儘可能多地在內存上添加內存。

// This starts at rptr->client_message and advances two indexes. 
rptr->client_message[2] = ...; 

的問題是有另一個零長度立即陣列rptr->client_message後，rptr->server_message。所以當你寫信給rptr->client_message時，你會覆蓋rptr->server_message。

讓我們簡化一下，你就會明白爲什麼。

struct region { 
    int len; 
    char ptr_client_message[0]; 
    char ptr_server_message[0]; 
}; 

以與您相同的方式初始化它。

size_t size = 4; 
struct region *rptr = malloc(sizeof(struct region) + size + size); 

現在我們似乎有兩個3字符串的地方。我們來添加一個。

rptr->ptr_server_message[0] = 'a'; 
rptr->ptr_server_message[1] = 'b'; 
rptr->ptr_server_message[2] = 'c'; 
rptr->ptr_server_message[3] = '\0'; 

printf("server_message: %s\n", rptr->ptr_server_message); 

這很好，它打印abc。現在我們來填充rptr->ptr_server_message。

rptr->ptr_client_message[0] = '1'; 
rptr->ptr_client_message[1] = '2'; 
rptr->ptr_client_message[2] = '3'; 
rptr->ptr_client_message[3] = '\0'; 

printf("client_message: %s\n", rptr->ptr_client_message); 

這很好，它打印123。那麼rptr->ptr_server_message？

printf("server_message: %s\n", rptr->ptr_server_message); 

也打印123！事實證明，他們指向相同的記憶。

// 0x7ff5f1404144 0x7ff5f1404144 
printf("%p, %p\n", rptr->ptr_client_message, rptr->ptr_server_message); 

因此，在一個結構中不能有兩個零長度的數組。