Cycle through List and Ordering - c

Question

I want to create a linked list.
The user adds numbers and the idea is that the numbers are inserted to the list in descending order.
Here goes what I did, but when rearranging, it just orders the first number...
int addInputNumber(numberList **node){
numberList *temp;
int userInput;
temp = (numberList*)malloc(sizeof(numberList));
//Memory Check
if ( temp == 0 )//out of memory, return 0
return 0;
//Get the users input
printf("Give me a Number!\n");
scanf("%d",&userInput);
//Add it to the list.
temp->numbero = userInput;
///Link to the list.
temp->next = *node;
*node = temp;
//Lets cycle through the list.
numberList *temp2;
int helpNumber;
temp2 = *node;
//Rearrange the list.
while(temp2 != 0){
if(temp->numbero < temp2->numbero){
//Switch position..
helpNumber= temp2->numbero;
temp2->numbero = temp->numbero;
temp->numbero = helpNumber;
temp2 = *node;// If we change number, we must cycle from the beginning AGAIN.
}//eof if
temp2 = temp2->next;
}//eof while
return 0;
}//eof addNUmber function.
Here's the structure just in case:
typedef struct dynamicNumberList {
int numbero;
struct dynamicNumberList *next;
}numberList;
I've got 2 quick questions.
Why would it only arrange the first number?
This list adds a space towards the left (visually), how could I make it so I can add a space to the right?

Answer 1

You need to get into the habit of creating one function per task, instead of cramming everything into a single one. It makes the code easier to read and reuse and reduces the chance of errors.
A correct implementation could look like this:
#include <stdio.h>
#include <stdlib.h>
typedef struct s_List
{
int n;
struct s_List *next;
} List;
void print_list (List *head)
{
List *ptr;
for (ptr = head; ptr; ptr = ptr->next) {
printf ("%d\t", ptr->n);
}
putchar ('\n');
}
List * make_node (int n, List *next)
{
List * node = malloc (sizeof(List));
node->n = n;
node->next = next;
return node;
}
void insert_number_front (List **head, int n)
{
*head = make_node (n, *head);
}
void insert_number_after (List *prev, int n)
{
prev->next = make_node (n, prev->next);
}
// If HEAD is sorted, it will stay sorted after insertion
void insert_number_sorted (List **head, int n)
{
List *ptr;
List *ptr2;
// search for the rightmost node whose number is smaller than n.
ptr2 = NULL;
for (ptr = *head; ptr; ptr = ptr->next) {
if (ptr->n >= n)
break;
ptr2 = ptr;
}
// If such a node exists we insert the new node after it,
// otherwise we insert it at the front of the list.
if (ptr2) {
insert_number_after (ptr2, n);
}
else {
insert_number_front (head, n);
}
}
int input_number ()
{
int n;
printf ("enter a number: ");
scanf ("%d", &n);
return n;
}
int main ()
{
List *head = NULL;
int i;
// By adding elements exclusively with insert_number_sorted()
// we ensure the list is always sorted
for (i = 0; i < 5; i++) {
int n;
n = input_number ();
insert_number_sorted (&head, n);
}
print_list (head);
return 0;
}
To answer your second question, what you have here is a singly linked list, which can be described by a pointer to the first node. If you want to be able to insert nodes at the back you need to maintain an additional pointer to the last node. However this is not necessary in this case.