Like Tree2Likes
  • 1 Post By smokeyangel
  • 1 Post By Salem

Linked list example with a pointer to pointer

This is a discussion on Linked list example with a pointer to pointer within the C Programming forums, part of the General Programming Boards category; Hi, I have a piece of code that is demonstrating a linked list. What I don't understand is what's happening ...

  1. #1
    Registered User
    Join Date
    Jun 2012
    Location
    Göteborg
    Posts
    16

    Linked list example with a pointer to pointer

    Hi,

    I have a piece of code that is demonstrating a linked list. What I don't understand is what's happening in the add function.

    There is a pointer to a pointer '**q'. Further down a pointer variable is assigned 'temp = *q'
    Why do I have to use '**q' as a parameter in the add function?

    My code is:
    Code:
    # include<stdio.h>
    # include<malloc.h>
    
    /* define the node */
    struct node {
    int         data;
    struct node *next;
    };
    
    /* Add a node to the linked list */
    void add(struct node **q, int num) {
      struct node *temp;
      temp = *q;
    
      /* first time there is no node therefore allocate */
      if(*q==NULL) {
        *q=malloc(sizeof(struct node));
        temp = *q;
      }
    
      /* add node after the last one */
      else {
        /* move on until the last node then create a new node*/
        while((temp->next)!=NULL) {
          temp = temp->next;
        }
        temp->next = malloc(sizeof(struct node));
        temp = temp->next;
      }
    
      // set data and pointer for the node, the last one always point at NULL
      temp->data = num;
      temp->next = NULL;
    }
    
    /* Display the linked list */
    void display(struct node *pt) {
    
      // traverse through all nodes until the last one
      while(pt!=NULL) {
        printf(" Data : %d\n",pt->data);
        printf(" next : %p\n",pt->next);
        pt = pt->next;
      }
    }
    
    /* ******* MAIN ******* */
    int main() {
    
      int          continue_add; // used if you want to continue adding elements
      int          your_choice;  // user choice of what to do
      int          num;          // user data
      struct node *ptr;          // always pointing at the first node after first have been added
    
      /* init values */
      ptr          = NULL;
      continue_add = 1;
    
      while (continue_add == 1) {
        printf(" Main Menu\n");
        printf("   1. Add element\n");
        printf("   Enter your choice ");
        scanf("%d",&your_choice);
    
        switch(your_choice) {
          case 1:
            printf(" Enter data to your node -> ");
            scanf("%d",&num);
            add(&ptr,num);
            display(ptr);
            break;
    
          default:
            printf(" Illegal choice! ");
        }// switch(your_choice)
    
        printf(" Do you want to continue (press 1 for yes) ");
        scanf("%d",&continue_add);
        printf("\n");
      }// while (continue_add==1)
    }// main()

  2. #2
    Registered User
    Join Date
    Mar 2010
    Posts
    535
    Quote Originally Posted by hzcodec View Post
    There is a pointer to a pointer '**q'. Further down a pointer variable is assigned 'temp = *q'
    Why do I have to use '**q' as a parameter in the add function?
    Because C passes parameters by value and you have this:

    Code:
      /* first time there is no node therefore allocate */
      if(*q==NULL) {
        *q=malloc(sizeof(struct node));
    If you just passed "ptr" then you'd have *q as an argument, which is just a pointer. If you then did
    Code:
        q=malloc(sizeof(struct node));
    You'd just write over the local parameter q. When you returned to main, ptr would still be NULL. Using a pointer to a pointer allows you to change the value of the actual pointer. It's just like passing a pointer to something else.

    All the rest of add() could function with just *q, it's just that initial write.
    Salem likes this.

  3. #3
    and the hat of wrongness Salem's Avatar
    Join Date
    Aug 2001
    Location
    The edge of the known universe
    Posts
    32,485
    Code:
    void foo ( int a ) {
      a = 10;
    }
    void bar ( int *a ) {
      *a = 10;
    }
    int main ( ) {
      int fred = 0;
      foo( fred );  // does this change fred?
      bar( &fred );  // what about this one?
    }
    If you've got a variable of type T in a function, and you want it to be modified by a called function, then you need to pass a pointer to it.
    That means the called function needs a T* parameter.

    So if your variable is already a pointer, this naturally leads to a pointer to a pointer.

    Look at line 17 in your code.
    AndiPersti likes this.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

Popular pages Recent additions subscribe to a feed

Similar Threads

  1. Linked List Pointer Seg Faults
    By DougD720 in forum C Programming
    Replies: 6
    Last Post: 07-25-2011, 09:24 PM
  2. Linked list using double pointer
    By linuxlover in forum C Programming
    Replies: 9
    Last Post: 03-15-2011, 02:53 PM
  3. next pointer in linked list
    By c_weed in forum C++ Programming
    Replies: 5
    Last Post: 11-15-2010, 05:42 PM
  4. Linked list pointer error
    By pattop in forum C Programming
    Replies: 1
    Last Post: 05-18-2004, 03:30 PM
  5. tail pointer - linked list
    By Space_Cowboy in forum C++ Programming
    Replies: 1
    Last Post: 12-02-2002, 02:05 AM

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21