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Help with linked list in C

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# 1  
Old 03-30-2011
Error Help with linked list in C
i have this code


Code:
    typedef struct client_list {
        char *client_name;
        struct client_list * next;
        int client_socket_fd;
    } client;
    client *current, *head; head = NULL;
    char *h="test";
    add_client(current, h, head, &client_socket_fd);
    printf(current->client_name);

when it prints, i get a segmentation fault, what could be wrong?

the function add_client is
Code:
void add_client(client *current, char *username, client *head, int *client_socket_fd) {
    current = (client *)malloc(sizeof(client)); 
    current->client_name = username;
    current->next = head;
    current->client_socket_fd = *client_socket_fd;
    head = current;
    current = head;
}

# 2  
Old 03-31-2011
I think the problem is, that you only work on the malloc'ed memory locally within the function.
You give add_client a pointer of type client. This is given by call by value, i.e. assigning the pointer a newly malloc'ed memory will not assign this memory to the pointer in the main function! Therefore your client *current is pointing to an undefined memory location even after call of add_client.

Better do this:
Code:
client *add_client(char *username, client *head, int *client_socket_fd) {
    client *current;
    current = (client *)malloc(sizeof(client)); 
    current->client_name = username;
    current->next = head;
    current->client_socket_fd = *client_socket_fd;
    return current;
}


Furthermore: What are you willing to achieve with:
head = current;
current = head
?
# 3  
Old 03-31-2011
ok i changed it to

Code:
    typedef struct client_list {         
        char *client_name;         
        struct client_list * next;         
        int client_socket_fd;     
    } client;
    client *current; current->next = NULL;
    current = add_client("test", current, &client_socket_fd);
    printf(current->client_name);

and still segmentation error...

Code:
client *add_client(char *username, client *head, int *client_socket_fd) {
    client *current = (client *)malloc(sizeof(client)); 
    current->client_name = username;
    current->next = head;
    current->client_socket_fd = *client_socket_fd;
    return current;
}

# 4  
Old 03-31-2011
Code:
client *current;

This is just a pointer. It doesn't point to anything yet. The very next thing you do is use this invalid pointer: current->next = NULL; That could be trying to write to any memory whatsoever.

Do current=NULL; instead. A NULL head will mean an empty list. If you follow the logic in your function, it still works fine when head is NULL -- 'next' will be set to NULL, meaning end of list, and it won't crash because nothing uses the contents of 'head'.
Last edited by Corona688; 03-31-2011 at 05:00 PM..
# 5  
Old 03-31-2011
if this makes a difference, the way i am doing the list is backwards, so when i want to add something to it, i create a new one in the function, then add the given one (which could be NULL or an actual list) to the next, then return the new one
# 6  
Old 03-31-2011
To be honest, I did not understand your last sentence.

Anyways it does not make much difference whether you want a new entry to be the first one (after the head) or the last one. It is all just rearranged the next pointer. In the first case you rearrange the pointer from the head element to your newly created one and from there to the previously first, in the latter case you just traverse the list and change the last pointer form NULL to the new entry, making sure that the pointer in the newly created element points to NULL
# 7  
Old 03-31-2011
i still don't follow what your saying....
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