Linked list code not terribly elegant - c

Question

This is a toy version of the code I really want to write:
#include <stdio.h>
#include <stdlib.h>
typedef struct node_s {
char val;
struct node_s *next;
} node;
int n;
char obtain_next_char(){
if (n<100) {
return 'a';
} else {
return 0;
}
}
int main() {
char c;
node *start, *curr;
n = 0;
start = malloc(sizeof(node));
curr = start;
while ((c=obtain_next_char()) != 0) {
curr->val = c;
curr->next = malloc(sizeof(node));
curr = curr->next;
}
curr->next = NULL;
printf("chain made, %d links\n", n);
curr = start;
while (curr != NULL) {
printf(">>>%c<<<\n",curr->val);
curr=curr->next;
}
}
First, regarding obtain_next_char, the function is highly simplified here. With the real function, there is no way to know ahead of time what the return value will be.
Now, the code works, but it results in an empty node at the end of the list, which I'm going to have to get rid of. Can this code be re-written to remedy this deficiency? I am trying to avoid duplicate code at the head of the loop, i.e. I don't want to set up the first node with code outside the loop and then set up the following nodes with the same code inside the loop.

Answer 1

node dummy, *start, *curr;
curr = &dummy;
while ((c=obtain_next_char()) != 0) {
curr->next = malloc(sizeof(node));
curr = curr->next;
curr->val = c;
}
curr->next = NULL;
start = dummy.next;

Answer 2

How about this:
start = NULL;
while ((c=obtain_next_char()) != 0) {
if (!start) {
start = malloc(sizeof(node));
curr = start;
} else {
curr->next = malloc(sizeof(node));
curr = curr->next;
}
curr->val = c;
curr->next = NULL;
}

Answer 3

I'm not sure what you want to achieve with this function because, here you are not scanning in any input :
char obtain_next_char(){
if (n<100) {
return 'a';
} else {
return 0;
}
}
maybe you need this :
char obtain_next_char()
{
char c;
scanf(" %c",&c); //scanning input
if (n<100) //if number of links < 100 then, return character
{
return c;
}
else //else return '0' (ascii character)
{
return '0';
}
}
but it results in an empty node at the end of the list
reason :
The problem is that a new node is being created at the end of while loop
while ((c=obtain_next_char()) != 0)
{
curr->val = c;
curr->next = malloc(sizeof(node)); //new node at end of each iteration
curr = curr->next;
}
So, when you end input, the last empty node remains
solution : (without first node with code outside the loop)
To avoid this, try creating node at the start of each iteration
while ((c=obtain_next_char()) != '0')
{
if(n != 0) //to avoid again creating for head node
{
curr->next = malloc(sizeof(node));
curr = curr->next;
}
curr->val = c;
n++; //increasing number of links
}
This way, you can avoid creating extra empty node at the end of each iteration.
I don't want to set up the first node with code outside the loop
while ((c=obtain_next_char()) != '0')
{
if(n != 0) //any other node
{
curr->next = malloc(sizeof(node));
curr = curr->next;
}
else //for head node
{
start = malloc(sizeof(node));
curr = start;
}
curr->val = c;
n++; //increasing number of links made
}
putting altogether your code would be :
#include <stdio.h>
#include <stdlib.h>
typedef struct node_s {
char val;
struct node_s *next;
} node;
int n;
char obtain_next_char()
{
char c;
scanf(" %c",&c);
if (n<100)
{
return c;
}
else
{
return '0';
}
}
int main() {
char c;
node *start=NULL, *curr=NULL;
n = 0;
while ((c=obtain_next_char()) != '0')
{
if(n != 0) //any other node
{
curr->next = malloc(sizeof(node));
curr = curr->next;
}
else //for head node
{
start = malloc(sizeof(node));
curr = start;
}
curr->val = c;
n++;
}
curr->next = NULL;
printf("chain made, %d links\n", n);
curr = start;
while (curr != NULL) {
printf(">>>%c<<<\n",curr->val);
curr=curr->next;
}
}
input :
1 2 3 4 5 0
output :
chain made, 5 links
>>>1<<<
>>>2<<<
>>>3<<<
>>>4<<<
>>>5<<<