Unexpected output in a linked list using C - singly-linked-list

Question

I've just started learning C and am fairly a beginner.
Today in school we learned linked list and I was able to put up a code...which thankfully is running without errors.
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *next;
}*head;//*temp;
void create(struct node **h,int num)
{
int i;
struct node *temp=*h;
for(i=0;;i++)
{
if(i>=num)
break;
temp->data=i;
temp->next=malloc(sizeof(struct node));
temp=temp->next;
}
temp->next=NULL;
}
void display(struct node **h)
{
struct node *temp=*h;
while(temp->next!=NULL)
{
printf("%d->",temp->data);
temp=temp->next;
}
printf("\b\b \b\b");
}
void append_end(struct node **h,int val)
{
struct node *temp=*h,*temp1;
//printf("'%d'",val);
while(temp->next!=NULL)
temp=temp->next;
temp1=malloc(sizeof(struct node));
temp1->data=val;
temp1->next=NULL;
temp->next=temp1;
}
void free_list(struct node **h)
{
struct node *temp=*h,*tail;
while(temp->next!=NULL)
{
tail=temp;
temp=temp->next;
free(tail);
}
h=NULL;
}
int main()
{
head=malloc(sizeof(struct node));
int i,num;
scanf("%d",&num);
create(&head,num);
//display(&head);
append_end(&head,5);
append_end(&head,6);
display(&head);
/*temp=head;
while(temp->next!=NULL)
temp=temp->next;
printf("%d",temp->data);*/
free_list(&head);
return 0;
}
The expected output should be
0->1->2->3->5->6
for input of 4
But instead I'm getting
0->1->2->3->(some garbage value)->5
I'll be glad if someone could point out my error(s), and/or link to any article that might help me in understanding the topic clearly.
Thanks in advance.

Answer 1

I've refactored your code. You can see it here: http://ideone.com/nZ55i
Your code was ugly, so I tweaked it into my own style. I suggest you find a style you like, that's easy for you to read.
Moved declaration of head into main method (used to be a global variable)
Functions that took pointer-to-pointer types for no reason have been modified to take pointers.
You used malloc(sizeof(struct node)) in at least 3 places, so I just up and made a function that does it for you.
Added a typedef to the declaration of the node struct, you can declare instances of it now with just node derp;
create function uses a for loop with an empty condition and a break, refactored to use a proper for loop with no break
The garbage value was caused by a bug in your create function that prevented it from writing the value to the last node it created. To fix it, I moved stuff around so it always assigns the data field but doesn't allocate a new node if it's at the end
Failure to write the last item in the list was caused by a bug in the display function that caused it to terminate before displaying the last node.
If this is a homework question and you turn in my code, I will find you, drive to your house, and shoot a rotten potato through your bedroom window with a cannon.

Answer 2

Okay, I started to put this in the comments, but putting code in comments will get brickbats :). This is not tested, and there are myriad ways to approach it (given I've diagnosed the problem correctly :) ), but one way that might work to fix your create method is offered as follows: (Note that I've simplified your loop, just using the normal termination condition for the 'for' loop construct, and eliminated the 'if..break' within the loop as its no longer needed. The "->next" member will always be initialized to a new member if there's one left to create, otherwise its NULL, so we don't need the ending assignment:
void create(struct node **h,int num)
{
int i;
struct node *temp=*h;
for(i=0;i<num;i++)
{
temp->data=i;
if (i==(num-1))
temp->next=NULL;
else
temp->next=malloc(sizeof(struct node));
temp=temp->next;
}
}