Question

I’ve read plenty of posts about this, but it seems, that my problem is a bit more specific. I think I would manage to allocate a dynamic 2d array.

Due to existing code, I have to use a typedef which is a static array. Now I want to store an unknown number of these arrays temporarily. I’ve tried several variations with pointers, but I can’t even compile it.

Following code should explain what I’m trying to do:

int                     iCount, i;
typedef unsigned char   Buffer[1024];

Buffer       *          BufferArray=NULL;

BufferArray = malloc(iCount * sizeof Buffer*);

for(i=0;i<iCount;i++)
{
    BufferArray[i] = malloc(sizeof(Buffer));
}

This is my version with fewest errors. The only one left is

error C2106: ‘=’ left operand must be l-value

I know this topic is tedious and occurred often enough. Though, I’m getting quite confused with the typedef, which is already a static array. So I thought, a ** pointer is not needed here.

Appreciate any help.

Answer 1

If the typedef confuses you, then remove it (in your head):

typedef unsigned char   Buffer[1024];  

So when we see Buffer, think of it as a unsigned char array of [1024].

Buffer * BufferArray=NULL; // really it's: unsigned char[] * 

Now you’re looking to hold onto an unknown number of these arrays. Well you’re not really using a 2D dynamic array, just a single dynamic array that happens to hold static arrays:

typedef unsigned char Buffer[1024];

int main(void) {
    Buffer * BA = NULL;
    int iCount = 5;
    BA = malloc(iCount * sizeof(Buffer));
    BA[0][0] = 10;
    return 0; 
} 

Now BA is a dynamically created array you can use to temporarily hold iCount number of static unsigned char arrays (Buffer).