在C99中的结构中将指针分配给2D空数组

Question

I have a 2d buffer that I would like to pass to a thread along with other values so I need to move it into a struct. When I try to do this, I cannot access the values in again in my struct and I need some help.

Here's my buffer:

void* mybuff[8][16384];

I would like to store a pointer to this 2D array in a struct like this:

typedef struct  {
  int i;
  void *mybuff;
} arg_struct_t;

However if I try to access the values:

    arg_struct_t *args = malloc(sizeof(arg_struct_t));
    args->i = index;
    args->mybuff = mybuff;

    fprintf(stderr, "%p\n", args->mybuff[0][0]);

I get an error:

main.c:104:47: warning: dereferencing ‘void *’ pointer [enabled by default]
           fprintf(stderr, "%p\n", args->mybuff[0][0]);
                                               ^
main.c:104:50: error: subscripted value is neither array nor pointer nor vector
           fprintf(stderr, "%p\n", args->mybuff[0][0]);
                                                  ^

I think I need to tell C99 more about the shape of the data I want to put into my struct, but I can't quite figure out the right way to do that. I appreciate any help or insights here.

Answer 1

The compiler doesn't know, what address in the memory it must calculate here:

args->mybuff[0][0]

because args->mybuff is just a pointer to void. To process a such expression the compiler needs info about 1) size of the elements of the array and 2) at least one of the dimensions of the array. Following code is more correct:

typedef void* TBuff[8][16834];
TBuff mybuff;

typedef struct  {
  int i;
  TBuff * mybuff;
} arg_struct_t;

arg_struct_t *args = malloc(sizeof(arg_struct_t));
args->i = index;
args->mybuff = &mybuff;
fprintf(stderr, "%p\n", (*args->mybuff)[0][0]);

Answer 2

At the risk of getting you farther over your head:

typedef struct  {
  int i;
  void* mybuff[8][16384];
} arg_struct_t;

But the general nature of void* is generic typing which is contrary to multidimensional arrays. This code looks strange as soon as that type appears and there's probably something else wrong.

Answer 3

Here is how to store the address of mybuff, and how to use that pointer later to access values.

int main()
{
    // Starting array    
    int* mybuff[8][16384];

    // Save the address of mybuff into 'p'. 'p' is a pointer to (int*)'s
    int** p = &mybuff[0][0];

    // Access the location of row=3, col=4 using 'p'
    int r = 3;
    int c = 4;
    *(p + (r*16384 + c)) = (int*) 123; // 123 is the value to assign

    return 0;
}

So you'll also need to pass the row and column sizes (8 and 16384) in your struct too in order to dereference the values. I've left out bounds checking in the example for brevity, but you should do that in real code.