Question: I have created a linked list struct in c 

```
struct node{
   int value;
   struct node* next;
};

```

a method to add a node at the start of the list :

```
void addFirst(struct node *list, int value){
    struct node *new_node = (struct node*) malloc (sizeof (struct node));
    new_node->value = value;
    new_node->next = list;
    list = new_node;
   }

```

I create a list (malloc and everything), then call this method, it adds the new node inside the method but when i get back to my main my old list remains unchanged. Using DDD debugger to check everything. How is this possible? I am not able to change the method signature so it has to be done like this.

Answer: 
If you really need to do it this way, you have to re-cast the pointer. Something like this:

```
struct node *my_list = null;
addFirst((struct node *)&my_list, 123);

void addFirst(struct node *list, int value){
    struct node **real_list = (struct node **)list;
    struct node *new_node = (struct node*) malloc (sizeof (struct node));
    new_node->value = value;
    new_node->next = *real_list;
    *real_list = new_node;
}

```
the node pointer could not be changed into the function in this way. in the function you are able to change the content of the pointer and not the address of the pointer. To do you have to pass your pointer of pointer `struct node **list`

here after how to do it:

```
void addFirst(struct node **list, int value){
    struct node *new_node = (struct node*) malloc (sizeof (struct node));
    new_node->value = value;
    new_node->next = *list;
    *list = new_node;
}

```

or you can do it in this way

```
struct node * addFirst(struct node *list, int value){
    struct node *new_node = (struct node*) malloc (sizeof (struct node));
    new_node->value = value;
    new_node->next = list;
    return new_node;
}

```

and in your cod you can get the head after the call of this function

```
head = addfirst(head,45);

```