Thread: 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.

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.