Swap items of void* pointer array without memcpy in C -

i writing school project, , need swap 2 items of void* pointer array. can 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 need swap items without memcpy, malloc, realloc , free. possible?

thank you

why don't swap in way?:

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

as qsort (swaps elements within 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;     (i = left + 1; <= 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); }