Question

I’m trying to make linked list similar too the one here:

linked list in C

That is to have the “head”, I called it first, inside another struct. However I found doing that change. Makes it hard to add values to the list_item struct. I have tried some few things to see if it works. It compiles, however when I run the code it will crash. Any help would be helpful here. I know the cause of the crash is when I want to point the new_node to the linked_list.

#include <iostream>

using namespace std;

struct list_item
{
    int key;
    int value;
    list_item *next;
};

struct list
{
    struct list_item *first;
};

int main()
{
    list *head;
    list *new_node;

    head = NULL;
    head->first = NULL;

    for(int i = 0; i < 10; i++)
    {
        //allocate memory for new_node
        new_node = (list*)malloc(sizeof(list));
        new_node->first = (list_item*)malloc(sizeof(list_item));
        //adding the values
        new_node->first->key = i;
        new_node->first->value = 10 + i;

        //point new_node to first;
        new_node->first->next = head->first;

        //point first to new_node;
        head->first = new_node->first;

    }

    //print
     list *travel;
     travel->first = head->first;

     int i = 0;
     while(travel != NULL)
     {
         cout << travel->first->value << endl;
         travel->first = travel->first->next;
     }

    return 0;
}

Answer 1

You are creating 10 lists, I think you might try to do something like this:

#include <iostream>

using namespace std;

struct list_item
{
    int key;
    int value;
    list_item *next;
};

struct list
{
    struct list_item *first;
};

int main()
{
    //Just one head is needed, you can also create this
    // on the stack just write:
    //list head;
    //head.first = NULL;
    list *head = (list*)malloc(sizeof(list));
    list_item *new_node = NULL;

    head->first = NULL;

    for(int i = 0; i < 10; i++)
    {
        //allocate memory for new_node
        new_node = (list_item*)malloc(sizeof(list_item));
        //adding the values
        new_node->key = i;
        new_node->value = 10 + i;

        //if the list is empty, the element you are inserting
        //doesn't have a next element

        new_node->next = head->first;

        //point first to new_node. This will result in a LIFO
        //(Last in First out) behaviour. You can see that when you 
        //compile
        head->first = new_node;

    }

     //print the list 
     list_item *travel;
     travel = head->first;

     while(travel != NULL)
     {
         cout << travel->value << endl;
         travel = travel->next;
     }

    //here it doesn't matter, but in general you should also make
    //sure to free the elements
    return 0;
}

This is what is going on. At first you only have one head and no elements.

head
  |
  |
  V
 NULL

Then you add your first element. Make sure that the “new_node->next==NULL”:

head
  |
  |
  V
node:   ------------------> NULL
key = 0
value = 10

Then you add another node in front but append your first node to its next node. you move the pointer from the head to the new node

head:
first
  |
  |
  V
node:   ---------> node:  -------------> NULL
key: 1             key: 0   
value: 11          value: 10  

etc.

Since you are using c++, you might consider using “new” and “delete”. Just replace

new_node = (list_item*)malloc(sizeof(list_item));

with

list *head = new list