confusion of hashtab and linked list

Code:

---

  struct nlist *lookup(char *);
  char *strdup(char *);

  /* install:  put (name, defn) in hashtab */
  struct nlist *install(char *name, char *defn)
  {
      struct nlist *np;
      unsigned hashval;

      if ((np = lookup(name)) == NULL) { /* not found */
          np = (struct nlist *) malloc(sizeof(*np));
          if (np == NULL || (np->name = strdup(name)) == NULL)
              return NULL;
          hashval = hash(name);
          np->next = hashtab[hashval];  /*******************/
          hashtab[hashval] = np;
      } else      /* already there */
          free((void *) np->defn);  /*free previous defn */
      if ((np->defn = strdup(defn)) == NULL)
          return NULL;
      return np;
  }

---

Please look at the line with comment of many stars.

In my understanding, np->next should point to the address of next structure in the linked list. Why here does it point back to the hashtab? I read this code again and again, but just can't understand it.

It's an example in K&R "The C Programming Language", 6.6, Table Lookup.
:confused:

I've figured out it. I misunderstood the expression.

It says, if hashtab[hashval] has a value already, np->next will be assigned with that value.
if hashtab[hashval] doesn't has a value, np->next will be assigned as NULL.

NULL is a value.

Hey, I too got confuse...
But I think that it might be like this:

/* the 'next' struct is pointed to 'hashtab[hashval]' */
np->next = hashtab[hashval];

/* 'hashtab[hashval] = np;' points to a new struct which is next in the list due to the pervious line*/
hashtab[hashval] = np;

If its wrong, pls correct me, I too want to know how it work.