C adding node to head of linked list

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 :

Answer 1

I have learned @Vlad Lazarenko answer,and I made the code like this ,is it right?

addFirst((struct node**)head,123);

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

Answer 2

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;
}

Answer 3

In C, if you want a function to be able to change a value it received in its arguments, you need to pass the address of that value. So, to change the value of the list pointer, you need to pass the address of the list pointer. Your addFirst() function should look like this :

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;
}

And when calling that function, you call it like this :

addFirst(&list, value);

Now, if you want to keep the function's signature, a possibility is to change the way you consider the head node. If you state that your head node is purpose is only to hold a pointer to the first value, but does not contain a value by itself, you can do something like this :

struct node *head;

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->next;
     list->next = new_node;
}

addFirst(head, 45);

Now, you only have all your functions that work on the list to be change so they work the same, to consider that 'head' is only pointing to the real first node of the list but is not a member of the list itself. The 'real' head is, for all practical purpose, head->next.

Answer 4

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.

Answer 5

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;
}

This is the correct code. the problem is that the poineter struct node *list You are passing can't be changed as it is a stack variable. If you change it to struct node **list you are passing a pointer to the first node of the list. now you can change that to point to the new first node of the list.

Answer 6

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);