Values added wont save in the node - c

Question

Please help values wont save once added into the list and nothing is printed as a result
#include <stdio.h>
#include <stdlib.h>
struct node{
char first[30];
char last[30];
char address[100];
char postal[6];
char number[10];
struct node* next;
};
typedef struct node node;
void print(node* root);
node* add(node* root);
void addinfo(node* root);
int main(){
node* root=NULL;
root=add(root);
print(root);
return 0;
}
Its supposed to add a new node at the end of the list then add values
node* add(node* root){
if(root==NULL){
root=(node*) malloc(sizeof(node));
addinfo(root);
root->next=NULL;
return root;
}
node* temp=root;
while(temp!=NULL){
temp=temp->next;
}
temp=(node*)malloc(sizeof(node));
addinfo(temp);
temp->next=NULL;
return root;
}
void print(node* root){
node* temp=root;
while(temp!=NULL){
printf("First Name: %s\n",temp->first);
printf("Last Name:%s\n",temp->last);
printf("Address:%s\n",temp->address);
printf("Postal%s\n",temp->postal);
printf("Phone Number:%s\n",temp->number);
temp=temp->next;
}
}
If I print in the function, the values print normally but once this functions ends all the values are gone.
void addinfo(node* root){
node* temp=root;
while(temp!=NULL){
temp=temp->next;
}
temp=malloc(sizeof(node));
printf("Please enter a name:");
gets(temp->first);
printf("Please enter a last name:");
gets(temp->last);
printf("Please enter an address:");
gets(temp->address);
printf("Please enter a postal code:");
gets(temp->postal);
printf("Please enter a phone number:");
gets(temp->number);
}

Answer 1

The simplest way of implementing what you want is keeping a pointer to the last item of a list:
typedef struct list {
node *first;
node *last;
} list;
Then you initialize a list with
list alist = { NULL, NULL };
and add items as follows:
void addNewNode(list *plist, node *pnode)
{
pnode->next = NULL;
if( plist->first == NULL)
plist->first = pnode;
else
plist->last->next = pnode;
plist->last = pnode;
}
fill the list:
while(there_is_something_to_read())
{
node *newnode = malloc(sizeof(node));
addInfo(newnode);
addNewNode(&alist, newnode);
}
note however, addInfo() should do just what its name says: add info to the node; no list manipulation:
void addinfo(node* pnode){
printf("Please enter a name:");
gets(pnode->first);
// .........
printf("Please enter a phone number:");
gets(pnode->number);
}
print it:
node *pnode;
for(pnode = alist.first; pnode != NULL; pnode = pnode->next)
printNode(pnode);
with
void printNode(node *pnode)
{
printf( "first name: %s\n", pnode->first);
//...
printf( "phone: %s\n", pnode->number);
printf( "\n");
}

Answer 2

The problem is in this code:
void addinfo(node* root){
node* temp=root;
while(temp!=NULL){
temp=temp->next;
}
temp=malloc(sizeof(node));
You carefully traverse the entire list until temp is NULL (which could be achieved more swiftly by writing temp = NULL; and then you overwrite the NULL with newly allocated memory, but don't hook the new node into the list.
This function cannot be used on an empty list. It can be revised to work on a non-empty list:
void addinfo(node *root)
{
if (root == NULL)
return; // Report error?
node *temp = root;
while (temp->next != NULL)
temp = temp->next;
temp->next = malloc(sizeof(node));
if (temp->next == NULL)
return; // Report error?
temp = temp->next;
temp->next = NULL;
…other code as before…
}