unable to print elements in a linked list in c

Question

I am just learning about linked list. While I am trying to insert elements in linked list I am unable to print those inserted elements.

int main()
{
    int i,x,n; 
    struct node* head = NULL; 
    printf("Enter number of elements to insert into likedlist :"); 
    scanf("%d",&n); 
    printf("Enter elements: "); 
    for(i=0;i<n;i++) 
    {
        scanf("%d",&x);
        insert(head,x);
    }
    print(head);
} 

struct node* insert(struct node* head,int x)
{
    struct node* p = (struct node*)malloc(sizeof(struct node));
    p->data = x;
    p->next = NULL;
    if(head == NULL)
    {
        head = p;
        return head;
    }
    p->next = head;
    head = p;
    return head;
}

here I am adding the elements to a linked list by changing it's head node(insert_front).

void print(struct node* n)
{
    while(n != NULL)
    {
        printf("%d -> ",n->data);
        n = n->next;
    }
    printf("NULL");
}

So, what's wrong with this code. Output is like this

Sample Input:

 Enter number of elements to insert into likedlist :5
 Enter elements: 1 2 3 4 5

Sample Output:

NULL

Compile with all warnings and debug info: `gcc -Wall -Wextra -g` with [GCC](http://gcc.gnu.org/). Use the debugger (e.g. [`gdb`](https://sourceware.org/gdb/current/onlinedocs/gdb/)) to understand the behavior of your program — Basile Starynkevitch, Apr 21 '18 at 06:02
StackOverflow is *not* a *do-my-homework* service. And you did not provide any [MCVE]. — Basile Starynkevitch, Apr 21 '18 at 06:03
Even with a `main` function, it is off topic here. Use a debugger, and read more about [linked lists](https://en.wikipedia.org/wiki/Linked_list). Remember that in C, argument passing is by value. — Basile Starynkevitch, Apr 21 '18 at 06:07
"I am just learning about linked list." That means it's homework. — Barmar, Apr 21 '18 at 06:10
[don't cast malloc](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — Barmar, Apr 21 '18 at 06:10
The code you've shown works fine. So the code you haven't shown must be broken. That's why the site rules require a [mcve]. The question is off-topic if it doesn't include a [mcve]. — user3386109, Apr 21 '18 at 06:11
My guess is you're not assigning the result of `insert()` back to `head`. — Barmar, Apr 21 '18 at 06:12
I can do this same program in java. But I am interested to learn how the functionality works in C. That's why I am asking "why I am unable to print linked list in C?" — Manoj Kare, Apr 21 '18 at 06:14
@ManojKare If you post the `main()` function we can tell you what you did wrong. — Barmar, Apr 21 '18 at 06:15
@ManojKare do you understand what the other commenters are saying? How are you calling 'insert()'? Show ALL relevant code! — Martin James, Apr 21 '18 at 06:17
FWIW, the `insert` function can be reduced to 4 lines. After the `malloc`, it's just `p->data = x; p->next = head; return p;` Note that assigning the `head` in the function just changes the local copy of `head`. It doesn't change the `head` in `main`. — user3386109, Apr 21 '18 at 06:18
int main() { int i,x,n; struct node* head = NULL; printf("Enter number of elements to insert into likedlist :"); scanf("%d",&n); printf("Enter elements: "); for(i=0;i — Manoj Kare, Apr 21 '18 at 06:29
@ManojKare By clicking the [edit] link under the question, you can add that code into the question itself, which makes it a lot easier to read. — user3386109, Apr 21 '18 at 06:30
@ManojKare edit it into the question, please, not in comments, and it looks like the other commenters were correct - you are calling insert() incorrectly. Those 'return head;' statements mean something! — Martin James, Apr 21 '18 at 06:31
Thank you! Now read @BasileStarynkevitch answer, which explains how to call insert() correctly. — Martin James, Apr 21 '18 at 06:40

Basile Starynkevitch · Accepted Answer · 2018-04-21T08:37:47.957

Read more about C programming (first some good tutorial book, then some reference site, and later refer to the C11 standard n1570). We cannot teach you it in a few paragraphs.

C uses a call-by-value evaluation strategy.

^{So, at least for newbies, it is conventionally recommended (but not required) to never use a formal argument as the left-side destination of some assignment, because any change to a formal argument is local to the function having that formal argument and does not impact the caller.}

So, Compile with all warnings and debug info: gcc -Wall -Wextra -g with GCC. Use the debugger (e.g. gdb) to understand the behavior of your program (your bug is probably not in the code chunk you show us).

^{Ability to understand the behavior of an entire program, and to debug it, is an essential skill for developers. Both the compiler's warnings and the debugger can assist you in understanding the behavior of a program. See also http://norvig.com/21-days.html for a useful insight.}

So the head formal argument in insert is a local copy of the actual argument from the caller.

In

if(head == NULL)
{
    head = p;
    return head;
}

you modify only that copy, not the original. So the head = p; above is completely useless (and confusing), you'll better just replace the block in braces above with simply return p;

We don't know how you call insert, and we cannot help more.

Perhaps insert should get the address of some pointer.... or perhaps your main should use the return value of it...

In your edited question, inside your main, replace

    insert(head,x); //WRONG

(which does not change the local head declared in main, even if insert is changing its first formal, because of call by value argument passing) with

    head = insert(head, x);

The tiny problem is believing (implicitly) that `head = p;` changes something on the calling site — Basile Starynkevitch, Apr 21 '18 at 06:15

unable to print elements in a linked list in c

1 Answers1

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