CODEWITHSUNDEEP
c
Mike
Mike

Reputation: 390

Dynamic memory allocation with realloc

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?

Upvotes: 2

Views: 53

Answers (1)

zwol
zwol

Reputation: 140445

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();
}

Upvotes: 2

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