different ways of adding node in linked list

Question

This simple code (adding element to linked list and printing it) works fine

#include <stdio.h>
struct node{
    int item;
    struct node* next;
};

void print_node (struct node* n){
    while (n!= NULL){
        printf("%d ", (*n).item);
        n = n->next;
    }
    printf("\n");
}

void append_node(struct node *list, struct node *n){

    while(list->next != NULL)
        list = list->next;

    list->next = n;

}

int main(){
    struct node n1, n2;
    n1.item = 1;
    n1.next = NULL;

    n2.item = 2;
    n2.next = NULL;

    print_node(&n1);
    print_node(&n2);

    append_node(&n1,&n2);

    print_node(&n1);

    printf("done\n");
    return 0;
}

If instead I define the append_node as following

void append_node(struct node *list, struct node n){

    while(list->next != NULL)
        list = list->next;

    list->next = &n;

}

and call it accordingly in the main (i.e., append_node(&n1, n2) ) I get a segmentation fault when running the program. And I don't understand why :)

struct node n is local variable. Invalid in the scope of the outside of the function. — BLUEPIXY
– BLUEPIXY, Commented Feb 19, 2016 at 9:20
At least half the LL questions asked in this tag have the same 'local var' issue:( — Martin James
– Martin James, Commented Feb 19, 2016 at 9:37
Ops, sorry, probably I did not search for the right keywords... BTW, it's mainly java's fault. ;) — Peter
– Peter, Commented Feb 19, 2016 at 9:39

Community · Accepted Answer · 2017-05-23 11:50:34Z

1

When you call append_node(struct node *list, struct node n), the argument n is copied on the function context.

When the function is leave, the context is freed, and the copy n of your data is lost.

You could use your function append_node(struct node *list, struct node n) if you make a copy of n (using malloc) before putting it in linked list.

EDIT

This may help you: What's the difference between passing by reference vs. passing by value?

edited May 23, 2017 at 11:50

CommunityBot

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answered Feb 19, 2016 at 9:20

Mathieu

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5 Comments

Peter Over a year ago

the content of n inside the function is erased from memory even though it has been linked to list?

Mathieu Over a year ago

Yes: in your linked list, you just put the address of n, which is local to the function. You may run add some printf("address of n is %p\n", &n); to visualise that. In that case you may notice that addresses generated by malloc and stack addresses are different.

Adeel Ahmed Over a year ago

All the function variables(arguments) are stored on stack and when function returns all its variables get destroyed(inaccessible, may contain some other data).

Peter Over a year ago

OK, then (just for the sake of understanding), I want to modify the function print_node so that it does not receive a pointer. I managed in writing it like this (it works): void print_node (struct node n){ while (1){ printf("%d ", n.item); if (n.next == NULL) break; n = *(n.next); } printf("\n"); } How could I write in the while something like n!=NULL, when n is not a pointer, in order to not do the break inside the loop?

jada12276 Over a year ago

@Peter: You can do so if you want. Just make sure to assign the next node pointers in a local struct node* n and using appropriate printf for it.

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