Loop for Linked Lists

Question

I really like this implementation of code but I don't want it to be hardcoded so can someone help me to make a loop for list = new_node(1); and so forth.

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

typedef struct node {
    int number;
    struct node *next;
} node;

node *new_node(int number)
{
    node *n = malloc(sizeof(*n));
    if (n == NULL)
    {
        perror("new_node");
        exit(1);
    }
    n->number = number;
    n->next = NULL;
    return n;
}

int main(void) {
    node *list;

    list = new_node(1);
    list->next = new_node(2);
    list->next->next = new_node(3);
    list->next->next->next = new_node(4);

    for (node *tmp = list; tmp != NULL; tmp = tmp->next)
    {
        printf("%i\n", tmp->number);
    }

    while (list != NULL) {
        node *tmp = list->next;
        free(list);
        list = tmp;
    }
    return 0;
}

Answer 1

As you are trying to add new nodes to the tail of the list then in this case it will be reasonable to define a two-sided singly-linked list.

For example

typedef struct node 
{
    int number;
    struct node *next;
} node;

typedef struct list
{
    struct node *head;
    struct node *tail;
} list;

In main you can declare a list like

list lst = { .head = NULL, .tail = NULL };

A function that appends nodes to the tail of the list can look like

node * new_node( int number )
{
    node *n = malloc( sizeof( *n ) );

    if ( n != NULL )
    {
        n->number = number;
        n->next = NULL;
    }

    return n;
}

int push_back( list *lst, int number )
{
    node *n = new_node( number );
    int success = n != NULL;

    if ( success )
    {
        if ( lst->tail == NULL )
        {
            lst->head = n;
        }
        else
        {
            lst->tail->next = n;
        }

        lst->tail = n;
    }

    return success;
} 

And in main the function is called like

push_back( &lst, 1 );

or

if ( !push_back( &lst, 1 ) )
{
    puts( "Error: there is no enough memory." );
}

Here is a demonstration program.

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

typedef struct node 
{
    int number;
    struct node *next;
} node;

typedef struct list
{
    struct node *head;
    struct node *tail;
} list;

node * new_node( int number )
{
    node *n = malloc( sizeof( *n ) );

    if ( n != NULL )
    {
        n->number = number;
        n->next = NULL;
    }

    return n;
}

int push_back( list *lst, int number )
{
    node *n = new_node( number );
    int success = n != NULL;

    if ( success )
    {
        if ( lst->tail == NULL )
        {
            lst->head = n;
        }
        else
        {
            lst->tail->next = n;
        }

        lst->tail = n;
    }

    return success;
} 

void display( const list *lst )
{
    for ( const node *current = lst->head; current != NULL; current = current->next )
    {
        printf( "%d -> ", current->number );
    }

    puts( "null" );
}

int main( void )
{
    list lst = { .head = NULL, .tail = NULL };
    enum { N = 10 };

    for ( int i = 0; i < N; i++ )
    {
        push_back( &lst, i );
    }

    display( &lst );
}

The program output is

0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

You will need to write yourself some other function as for example a function that will free all the memory allocated for the list.