**linked list

i have a question, i have an assignment that im supposed to do involving a linked list, i've looked into them a little and have some knowledge, but my prof. wants me to use a pointer to a pointer??? im supposed to write a function w/a prototype like this

void BuildPhoneBook(PhoneBook **headPP, const char *str)

the 1st parameter is (im assuming) a pointer to a pointer to the 1st node in the linked list, the second is the info that will go in that particular node. im clueless where to start...

NO!, im not asking you to do my homework, i just need a shove in the right directioin

Well, how well do you understand pointers? Jumping into linked lists without a good understanding of pointers is like doing calculus with no algebra background.

Here's a quick tutorial. I'll assume that you know a pointer is just an address value and what it means to dereference a pointer. Based on that, all you really need to know is that all function parameters in C are "passed by value". That means that any changes that a function makes to parameter vars are not reflected in the calling parameter. For example

Code:

---

void foo(int n)
{
    n = 5;
}//foo

int main()
{
    int i = 0;

    foo(i);

    printf("%d",i);
    //Output: "0"

    return 0;
}//main

---

Here, foo() does not change the value of i.
Ok, so maybe you knew all this to begin with. Well, lets extend this to pointers...

Code:

---

void foo(int *n)
{
    n = NULL;
}//foo

int main()
{
    int i = 0;
    int *p = &i;
   
    printf("address before: %x\n",p);
   
    foo(p);

    printf("address after: %x\n",p);

    //Output: <the address's will be the same>

    return 0;
}//main

---

As you can see, the pointer is passed by value, therefore any changes done in foo() are not reflected in main().
So what if we really want the function to make a change to a parameter that is seen outside the function? The awnser is to pass in a pointer to whatever it is you want the function to modify. So to modify our original foo()....

Code:

---

void foo(int *n)
{
    *n = 5;
}//foo

int main()
{
    int i = 0;
   
    foo(&i);

    printf("%d",i);
    //Output: "5"

    return 0;
}//main

---

Here, foo() was able to modify the value of i in main(). This is called pass by reference (via a pointer in C). So to extend the concept to pointers, if you want to pass a pointer by reference then you have to have a pointer to a pointer:

Code:

---

void foo(int **n)
{
    *n = NULL;
}//foo

int main()
{
    int i = 0;
    int *p = &i;
   
    printf("address before: %x\n",p);
   
    foo(&p);

    printf("address after: %x\n",p);
    //Output: <address after will be 0>

    return 0;
}//main

---

Phew.....so why have "PhoneBook **headPP"? So you can modify the value of headPP within BuildPhoneBook().
If you didn't learn (or remember) anything, maybe I put someone else's learn on ;)

gg

ok, here's what i have so far, im not sure if this is correct???

Code:

---

void BuildPhoneBook(PhoneBook **headPP, const char *str)
{
        PhoneBook *header = headPP; //first node in list
       
        PhoneBook *conductor; //used to point to each node
       
        header = new PhoneBook;
       
        header->next = NULL;

        conductor = headPP;

        if(conductor != NULL)
        {
                while(conductor->next != NULL)
                {
                        conductor = conductor->next; 
            }
        }

}

---

maybe im just going about this the wrong way, but the only way i've seen linked lists done is w/the single pointer???

Revisit pointers in your class book/notes and study my tutorial. Notice that foo(int **n) had "*n = " somewhere in it, but your function does not have "*headPP = " anywhere in it.

Also, you need to understand things like:
- how is an empty list represented?
- when I add a new node to the list, do I add to the end of the list or the begginning? (hint: "**headPP" awnsers this one)

gg

im still kinda lost, i've reviewed what you posted and i understand passing by referecne(i think) but im not seeing what you mean?

Code:

---

void BuildPhoneBook(PhoneBook **headPP, const char *str)
{
        *headPP = new PhoneBook;

        PhoneBook *conductor;

        conductor = headPP;

        if(conductor != NULL)
        {
                while(conductor->next != NULL)
                {
                        conductor = conductor->next; 
            }
        }

}

---

assuming the same as codeplug check this link out :p

Linked List

You're getting closer, but you need to really think about and understand every single line that you write. For example, what were you thinking <don't read sarcastically> when you wrote:

Code:

---

conductor = headPP;

---

conductor is of type PhoneBook* and headPP is of type PhoneBook** - why whould you assign vars of differing "type" like that?

The only reason headPP is passed in is so that the caller of the function has a pointer to the list once you are done building it. Starting thinking about actually building the list and then worry about "*headPP = <something>" later.

Think along these lines:

Code:

---

while (not done building the list)
{
  create a new node to add to the list
  add the new node to the list by prepending it to the list (newnode->next = head; head = newnode)
}

---

Don't make me do your homework, cuz I will :)

gg

nah, i wont, thanks for all the help though, sorry im not picking this up real quick(not really in the programming mood today) im slowly gettin it though, i fixed my initializer thing....anyway, back to work

Keep it up. I can't stress enough how important it is to really understand pointer semantics like the back of your hand. Once you have that down, applying what you know is alot easier. Read up on that link that whistlenm1 posted, thats good stuff. Decompose the code line by line and understand it. Post here if there is anything you don't understand or need clarification on.

gg