C++ Linked list - c++

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

Answer 2

The next line only allocates memory for your list struct. The list contains only a pointer, you must also allocate memory for new_node->first before assigning to any of its members.
//allocate memory for new_node
new_node = (list*)malloc(sizeof(list));

Answer 3

I think you want something more like this:
#include <iostream>
#include <cstdlib>
using namespace std;
typedef struct tag_list_item
{
int key;
int value;
struct tag_list_item *next;
} list_item;
typedef struct
{
list_item *head;
} list;
int main()
{
list my_list;
list_item *new_node;
list_item *previous_node = NULL;
my_list.head = 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(previous_node == NULL)
{
my_list.head = new_node;
}
else
{
previous_node->next = new_node;
}
previous_node = new_node;
}
//print
list_item *iter = my_list.head;
while(iter != NULL)
{
cout << iter->value << endl;
iter = iter->next;
}
return 0;
}
Changes of note:
For malloc, I added:
#include <cstdlib>
I changed your list structures to typedefs, had to declare "next" using the tag since the typedef isn't complete at that point
typedef struct tag_list_item
{
int key;
int value;
struct tag_list_item *next;
} list_item;
I changed your list name to "my_list" and declared it directly (without the pointer). In this case you can just have the compiler allocate it automatically on the stack.
list my_list;
I keep a pointer for "previous_node" so that you can assign the "next" pointer much more easily. Notice that the first node allocated is pointed to by the "head" pointer in the list structure. I believe that is the traditional name for the pointer to the first element in a list.
if(previous_node == NULL)
{
my_list.head = new_node;
}
else
{
previous_node->next = new_node;
}
previous_node = new_node;

Answer 4

head = NULL;
head->first = NULL;
There's the issue. You can't follow a pointer and set it to NULL if you've set the pointer itself to NULL.
That should be
head = malloc(sizeof(list));
head->first = NULL;
That should fix your code.
Hope that helps,
Billy3
EDIT: There's also an issue with your FOR loop. When you allocate the list, you should only allocate the list itself once. When you insert an item, you only allocate a list_item. You're assigning a list pointer to a member which accepts only a list_item pointer ;)
See Gabe's post for a demonstration of correct behavior :)

Answer 5

Look at your struct declaration
struct list_item
{
int key;
int value;
list_item *next;
};
That should be
struct list_item
{
int key;
int value;
struct list_item *next;
};
Hope this helps,
Best regards,
Tom