Storing data Using Linked List - c

Question

I am trying to write a c program that ask user some information about a brand and store it into linked list. But whenever user enters data it always overwrites the previous data, instead I wanted it to create new node and store it there. This is what I manage so far. How can I do it without using any temporary nodes.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int count = 1;
struct modelNode{
char name[50];
int year;
int amount;
struct modelNode *next;
};
struct modelNode * addModel(struct modelNode *p);
void getBestModel(struct modelNode *p);
int main(){
int command;
struct modelNode * modelList = NULL;
do
{
printf("1. Add a model\n");
printf("2. Display the model with the highest selling amount\n");
printf("3. Exit\n");
printf("Enter command: ");
scanf("%d", &command);
switch(command)
{
case 1:
addModel(modelList);
printf("test %s\n", modelList->name);
//printf("test %s\n", modelList->next->name);
break;
case 2:
printf("45\n");
break;
case 3:
puts("Bye");
break;
default:
printf("default\n");
}
}while(command != 3);
return 0;
}
struct modelNode * addModel(struct modelNode *p){
int iter = 0;
while(iter<count)
{
p = (struct modelNode*)malloc(sizeof(struct modelNode));
p->next=NULL;
iter++;
}
printf("test %s\n", p->name);
printf("Enter the name: ");
scanf("%s", &p->name);
printf("Enter the release year: ");
scanf("%d", &p->year);
printf("Enter the selling amount: ");
scanf("%d", &p->amount);
p->next=NULL;
printf("test %s\n", p->name);
count++;
}
}

Answer 1

This looks like a school exercise, so i doubt anyone here will solve it for you.
But you should try to insert "at the head" , you don't need temp nodes.
Pass the [head] as a param of function addNodes:
Inside addNodes:
Create [new-node]
Set the "next" pointer from [new-node] to the [head]:
[new-node]->next = [head]
Return [new-node]
Resulting linked list (head is the old-head) :
[new-node] ----> [head]
If you do this twice you'll get this:
[newer-node] -----> [new-node] ----> [head]
And so on....
If you want to insert as "tail" , you just need to traverse the linked list recursively, this way you don't use temp nodes.
If i missed the point, please leave a comment explaining further.

Answer 2

The generic functions for adding a node to a linked list would be
typedef struct node_t {
struct node_t* next;
} node_t;
node_t* add(node_t* list, node_t* new_node) {
if (list == NULL) {
return new_node;
} else {
node_t* tmp = list;
while (tmp->next != NULL) {
tmp = tmp->next;
}
tmp->next = new_node;
return list;
}
}
and you could use it like
node_t* list = NULL;
list = add(list, malloc(sizeof(node_t)));
list = add(list, malloc(sizeof(node_t)));
list = add(list, malloc(sizeof(node_t)));
list = add(list, malloc(sizeof(node_t)));
// list would now be a list of 4 nodes
and if you really don't want to use a tmp variable to walk the list, you can do it like this
void add_without_tmp(node_t* list, node_t* new_node) {
if (list == NULL) {
// some error handling
} else {
while(list->next != NULL) {
list = list->next;
}
list->next = new_node;
}
}
node_t* list = malloc(sizeof(node_t));
add_without_tmp(list, malloc(sizeof(node_t)));
add_without_tmp(list, malloc(sizeof(node_t)));
add_without_tmp(list, malloc(sizeof(node_t)));
// list would now be a list of 4 nodes
Edit: fixed the error with the declaration of node_t and added the missing * to the statement node_t* tmp = list;.