Swap items of void* pointer array without memcpy in C

Question

I am writing some school project, and I need to swap two items of void* pointer array. I can do this with something like following code:

void swap(void *base, int len, int width)
{
    void *p = malloc(width);

    memcpy(p,base,width);
    memcpy(base,(char*)base+width,width);
    memcpy((char*)base+width,p,width);

    free(p);
}

But I need to swap items WITHOUT memcpy, only with malloc, realloc and free. Is that even possible?

Thank you

Instead of copying the memory, why dont you just swap the addresses? — huseyin tugrul buyukisik, May 14 '13 at 12:26
I have tried that, but I think it is just impossible with void* array... If it is possible, can you post some code here, please? — Firzen, May 14 '13 at 12:28
void is your native length-unsigned int. Just cast it or use it as a pointer-arithmetic to swap. You dont lose those variables unless you use free(). Or you lose if you use smart-pointers? — huseyin tugrul buyukisik, May 14 '13 at 12:28

David Ranieri · Accepted Answer · 2013-05-14T12:39:35.237

2

Why don't swap in this way?:

void swap(void *v[], int i, int j)
{
    void *temp;

    temp = v[i];
    v[i] = v[j];
    v[j] = temp;
}

As qsort does (swaps elements within the array):

void sort(void *v[], int left, int right, int (*comp)(const void *, const void *))
{
    int i, last;

    if (left >= right) return;
    swap(v, left, (left + right) / 2);
    last = left;
    for (i = left + 1; i <= right; i++) {
        if ((*comp)(v[i], v[left]) < 0)
            swap(v, ++last, i);
    }
    swap(v, left, last);
    sort(v, left, last - 1, comp);
    sort(v, last + 1, right, comp);
}

edited May 14 '13 at 12:39

answered May 14 '13 at 12:29

David Ranieri

39,972
7
52
94

Yes, that is it, thank you very much. I was too sticked to teacher's words about free, malloc and realloc. – Firzen May 14 '13 at 12:52

score -1 · Answer 2 · answered May 14 '13 at 12:39

The array contents can be swapped in place, using only a char as a temporary variable.

void swap(void *base, int len, int width)
{
  int i;
  char t;

  for (i = 0; i < width; i++)
  {
    t = base[i];
    base[i] = base[i + width];
    base[i + width] = t;
  }
}

Swap items of void* pointer array without memcpy in C

2 Answers2

Linked