哈希表中指针的指针

Question

i don't quite understand why the following code does what it does.

void initializeTable()
{
 NodeT* (*hashTable) ;
 *hashTable=(NodeT*)malloc(30*sizeof(NodeT));
  int i;
  for(i=0;i<30;i++)
    {
      (*hashTable)[i].info=(char*)malloc(10*sizeof(char));
       strcpy((*hashTable)[i].info,"a");
        cout<<(*hashTable)[i].info<<" ";

    }
}

I am trying to understand how hash Tables work. In the first version of my program I used something like NodeT* hashTable[arraySize] in order to create an array of pointers to NodeT , in order to do the chaining. After this I decided to try and allocate dinamycally memory for the array, so I tried to use the (I think) equivalent NodeT* (*mockTable) .

My question is the following: why do I have to use (*hashTable)[i].info , and not (*hashTable)[i]->info ? As far as i concerned, hashTable is a pointer which points to an array of pointers to NodeT , so this shouldn't happen. What am i getting wrong here?

Answer 1

You got the basic idea but it seem you miscount the pointers your hashTable is a pointer to pointer (**), and when you access it in the strcpy you actually get into the value of this double pointer the 1'st is with the array index [i] and the second is with the *hashTable which is when accessing a variable calling the Value inside it. so if you define a duel pointer variable StrutT **temp; and you want to access a field inside it you can *temp[i].field but in case you would only want to access at level of a single pointer (same **temp) you would need: temp[i]->field or *temp->field which both only "pop" the top most pointer