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: 
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);

```
All is good, but in the `void addFirst(struct node *list, int value)` function the `list` is passed by value. Which means that a pointer is being copied and assignment of a new address to that pointer inside `addFirst` function is not visible to the caller of `addFirst`. To solve it, you have to either pass a pointer by pointer (`struct node **`) or make it a return value and require a caller to use it as a new "head".

And don't forget `;` after a struct declaration.