Thread: Question on free() ing a hash table

Excercise 6-5

Write a functon undef that will remove a name and defintion from the table maintained by lookup and install.

Code:

unsigned hash(char *s);

void undef(char *s)
{
 int h;
 struct nlist *prev, *np;

 prev = NULL;
 h = hash(s);
 for(np = hashtab[h]; np!=NULL; np = np->next){
       if(strcmp(s, np->name) == 0)
            break;
       prev = np;
 }

 if(np !=NULL) {
   if(prev == NULL)
      hashtab[h] = np->next;
   else
      prev->next = np->next;
  free((void *) np->name);
  free((void *) np->defn);
  free((void *) np);
 }
}

The question is, how come I have to free() both np->name and np->defn? Ie, how come I can't just free() np?

The question is, how come I have to free() both np->name and np->defn? Ie, how come I can't just free() np?

I assume our data structure would look some thing like this.

Code:

struct nlist 
{
       char *name;
       char *defn;
};

Once your create the node and allocated memory for name and defn, these two pointer are now pointing at the newly allocated memory location. And I belive you have should be allocating somewhere in the code. Have a look at your. Hence in order to free these allocated memory as well you free name and defn as well. Just freeing np would normally lead up to memory leak, which not you wanted.

If you had something like this

Code:

struct nlist 
{
       char name[10];
       char defn[10];
};

Then you could have just freed np.

ssharish