How to return lists in c functions?

Question

As the title says I need to write a function which fill a list with int typed by the user, and later on print them, but i'm having troubles with the insert function, which does not actually put the values typed in the list. Here's the code:

the typedefs:

typedef struct list_element {
    int value;
    struct list_element *next;
} item;

typedef item *list;

and the functions:

list lins(list l)
{
    int i;
    list root = NULL;
    printf("inserire dati ('0' per terminare):\n");
    do
    {
        scanf("%d", &i);
        l = (list)malloc(sizeof(item)); 
        l->value = i;
        l->next = root;
        root = l;
    } while (i != 0);
    return l;
}

void showlist(list l)
{
    printf("lista completa:\n");
    while (l != NULL)
    {
        printf("%d ", l->value);
        l = l->next;
    }
    printf("\n");
}

Sorry if my code is poorly written but I'm having a hard time understanding the concept of lists.

Edit: added functions calls

main()
{
    lins(l);
    showlist(l);
}

Edit 2: here's the output of the code:

output

Edit 3: Here's the sample of code my professor give us to work with lists:

http://www.mediafire.com/file/wj152pw1ojkxf1j/10a_-Liste-_2018.pdf

Answer 1

Your insertion function inserts in reverse order. That is because you assign root to l->next and l to root. Also the number 0 is included in the list when the user tries to end the insertion process.

In main(), you don't seem to assign the return value of the function lins() to a variable, and if you passed l declared as List l = NULL; to the function lins(), then a copy of it will be used, and so, the original variable won't be altered. What you really need is something like:

#include <stdio.h>
#include <stdlib.h>

typedef struct list_element {
    int value;
    struct list_element *next;
} item;

typedef item *list;


list lins(void)
{
    list node = NULL;
    list root = NULL;
    list tail = NULL;

    printf("inserire dati ('0' per terminare):\n");

    int i;
    scanf("%d", &i);
    while(i != 0)
    {
        node = (list)malloc(sizeof(item));
        node->value = i;
        node->next = NULL;

        if (!root)
            root = node;
        else
            tail->next = node;
        tail = node;
        scanf("%d", &i);
    }
    return root;
}

void showlist(list l)
{
    printf("lista completa:\n");
    while (l != NULL)
    {
        printf("%d ", l->value);
        l = l->next;
    }
    printf("\n");
}


int main(void)
{
    list head = lins();
    showlist(head);

    system("pause");
    return 0;
}

Of course, I should point out that you should free the list once you are done with it.

Answer 2

First of, I see no point in passing item *l as an argument to lins function.

The root or better, head of the list should not be changed once you assign something to it, unless you want to delete the list.

void lins(item **head){
    item *new_node = malloc(sizeof(item));
    int i;

    if(!new_node){
        printf("Memory allocation failed!\n");
        exit(1);
    }

    scanf("%d", &i);

    new_node->value = i;
    new_node->next = NULL;

    node *temp = *head;

    if(!temp)
        *head = new_node;
    else{
        while (temp->next != NULL){
            temp = temp->next;
        }
        temp->next = new_node;
    }
    return ;
}

After you write lins like this, showlist can look as follow:

void showlist(item *head){
    printf("List is: \n");
    do{
        printf("%d", head->value);
        head=head->next;
    }while(head!=NULL);
}

As for the main function you need to declare the head first.

int main(){
    item *head = NULL;
    lins(&head);
    showlist(head);
    return 0;
}

more detailed explanation to follow...

Answer 3

So from what I understand, you want to prepend to your linked-list. I will try to use your typedefs, though many comments have given better options to use, you could change the implementation later.

Realise that lins(list l) is never utilising the value of the parameter l passed to it. That is overriden by the malloc. It is prepending elements to the list, and finally returns the new head of your list. That new head has to be passed to showlist() for printing. What you might be doing is initialising one element of the list in main, and passing that to lins and showlist. lins doesnt bother with the parameter and returns a completely new list, but when you pass that old list to showlist it is interpreted as an empty list.

So you can simply change usage of lins to be used as a creator of linked-lists.

So declaring list lins() {...} and using it in main() like this should work:

int main() {
    list l;
    l = lins();
    showlist(l);
    return 0;
}

Answer 4

struct list_element {
int value;
struct list_element *next;};
typedef struct list_element * list;

void lins(list *l){
list tmp = NULL;
list last = *l;
int i;
if (last){
    while(last->next){ // go to last child of list
        last = last->next;
    }
}else{
    last = (list)malloc(sizeof(struct list_element));
}

printf("inserire dati ('0' per terminare):\n");
do
{
    scanf("%d", &i);
    last->value = i; 
    last->next = (list)malloc(sizeof(struct list_element));
    last = last->next;
} while (i != 0);}

void show(list li){
list tmp = li;
if(li){
    while(tmp){
    printf("%d at %p\n",tmp->value, tmp);
    tmp = tmp->next;
    }
}}

int main(){
list a = (list)malloc(sizeof(struct list_element));
lins(&a);
show(a);
return 0;}