Printing a linked list of integers in C - c

Question

I'm trying to get an integer of any size to a linked list in C. But when I print the list, a zero is printed always after the the integer.
Please note that I'm adding each digit of the integer to the head.(head has the 0th place of the integer)
#include <stdio.h>
#include <stdlib.h>
struct node
{
int digit;
struct node* next;
};
void get_number(struct node** head);
int create_node(int digit, struct node** head);
void printlist(struct node* head);
int main()
{
struct node* head1 = malloc(sizeof(struct node*));
get_number(&head1);
printlist(head1);
return 0;
}
int create_node(int digit, struct node** head)
{
struct node* tmp = malloc(sizeof(struct node*));
tmp -> digit = digit;
tmp -> next = *head;
*head = tmp;
}
void printlist(struct node* head)
{
struct node* curr = head;
if(!head)
return;
while(curr != NULL )
{
printf("%d",curr -> digit);
curr = curr -> next;
}
}
void get_number(struct node** head)
{
int k;
char c;
c = getchar();
while(c != '\n' && c != ' ')
{
k = c - '0';
create_node(k, head);
c = getchar();
}
}
when I input 123456, the output is 1234560.
I tried to find solution, but couldn't. Please help

Answer 1

You are one more node than necessary when you allocate to head1. You simply need to call the function get_number() as:
struct node* head1 = 0;
get_number(&head1);
which would set the next of the last element (i.e. the first allocate node) to 0 and rest of the logic would be fine.
You also need to correctly call malloc() and change the type of c to int (to handle EOF) as noted in the comments. My preferred way is to allocate memory is:
TYPE *p = malloc(sizeof *p);