Question

Let’s say I’ve calloc’d some memory for myself:

byte *header = calloc(5 + ZHEADERSIZE, sizeof(byte));

This gives me an array of 0’s. But what if I want to make a custom initialization of the memory? The following code is a fairly detailed initialization for an array literal. Note that it actually contains variables, so I can’t just memcpy all of the array over. I’m wondering if I can replicate this style of initialization for a block of memory that is malloc’d out without having to insert them manually.

unsigned char zhead[] =
{
    0x00, 37, 0, 218, 0xFF,

    0x50, 0x4b, 0x03, 0x04,
    0x14,
    0x00,
    0x00, 0x00,
    0x08, 0x00,
    0x08, 0x03,
    0x64, 0x3c,
    0xAA, 0xBB, 0xCC, 0xDD,
    csize, (csize>>8), 0, 0,
    uncsize, (uncsize>>8), 0, 0,
    0x07, 0x00,
    0x00, 0x00,
    'r', '/', 'r', '.', 'z', 'i', 'p'
};

Answer 1

You cannot initialize dynamically allocated memory in C other than to zero using calloc. You can however provide a static prototype array from which you copy the data:

const char prototype[] = { 'a', 'b', 'c' };

int main()
{
   char * data = malloc(sizeof prototype);
   memcpy(data, prototype, sizeof prototype);

   data[1] = 'z'; // fill in custom data
}