예제 #1
0
파일: hash_newtab.c 프로젝트: Dasudian/otp
ei_hash *ei_hash_newtab(int tabsize)
{
  ei_hash *tab=NULL;
  int bucketpos=sizeof(*tab);
  
  /* make sure size is odd, then increase until prime */
  tabsize |= 0x1; 
  while (!ei_isprime(tabsize)) tabsize +=2;

  /* we will only do one malloc, so "sizeof(*tab)" 
   * must be adjusted to align tab->tab properly
   */
  ei_align(bucketpos);

  /* single malloc, then fill in all fields */
  if ((tab = malloc(bucketpos + (tabsize * (sizeof(*(tab->tab))))))) {
    tab->tab = (ei_bucket **)((char *)tab + bucketpos);
    memset(tab->tab,0,tabsize*sizeof(*(tab->tab)));
    tab->hash = ei_dohash;
    tab->size = tabsize;
    tab->npos = 0;
    tab->nelem = 0;
    tab->freelist = NULL;
  }

  return tab;
}
예제 #2
0
/* move the elements from oldtab to a new table newsize. The old table
 * is freed and the caller should discard the pointer. On failure
 * (i.e. if malloc fails) return the old table and do nothing. 
*/
ei_hash *ei_hash_resize(ei_hash *oldtab, int newsize)
{
  ei_hash *newtab=NULL;
  ei_bucket *b, *next;
  int i,h;

  /* make sure size is odd, then increase until prime */
  newsize |= 0x1; 
  while (!ei_isprime(newsize)) newsize +=2;

  if (newsize == oldtab->size) return oldtab;
  
  /* make a new table */
  if (!(newtab = ei_hash_newtab(newsize))) return oldtab;
  newtab->hash = oldtab->hash;

  /* move the buckets, rehashing */
  /* note that this will reverse the order of any chains */
  for (i=0; i<oldtab->size; i++) {
    b=oldtab->tab[i];
    while (b) {
      next = b->next;
      h = b->rawhash % newtab->size;
      b->next=newtab->tab[h];
      if (!newtab->tab[h]) newtab->npos++;
      newtab->tab[h]=b;
      b = next;
    }
  }
  /* the new table has the same number of elements as the old one */
  newtab->nelem = oldtab->nelem;

  /* the new table takes over the freelist from the old one */
  newtab->freelist = oldtab->freelist;

  /* now it's safe to remove the old table */
  free(oldtab);

  return newtab;
}