Suppose we have this code:
int *h; for(int i=0;i<5;i++){ h = malloc(sizeof(int)); h[i] = i; } The issue I have here is that I want to start with an empty array, i.e. just declaringint *h, and then *h will grow at the runtime using realloc. I tried using this example but it does not allocate a sequential memory places and also does not work. I understand that realloc works after allocating by malloc so is there any workaround of that?
To accommodate exactly what you are trying to do, realloc(NULL, size) does the same thing as malloc(size). You can write your loop like this:
int *h = 0; size_t avail = 0; for (size_t i = 0; more_data_available(); i++) { if ((i+1) * sizeof(int) > avail) { avail = avail == 0 ? 8*sizeof(int) : avail*2; int *nh = realloc(h, avail); if (!nh) abort(); h = nh; } h[i] = next_data_item(); } But note the convoluted ?: expression I had to use to enlarge avail. That would be cleaner if I start the loop with some space already allocated; and then I can just use malloc normally:
size_t avail = 8 * sizeof(int); int *h = malloc(avail); if (!h) abort(); for (size_t i = 0; more_data_available(); i++) { if ((i+1) * sizeof(int) > avail) { avail *= 2; int *nh = realloc(h, avail); if (!nh) abort(); h = nh; } h[i] = next_data_item(); } 
i for which h[i] is meaningful data. (This is not the same as avail in my sample code.)
h = malloc(sizeof(int));allocates only aninth[0]and leaking the previoush