Passing linked list that consists of three structures to a function

Question

I know how create a linked list with two structures

To do so, I declare a structure that contains all the necessary data. It looks like this:

struct Data{
    int numb;
    int date;
}

Second structure represents a node that has a head (ie first element of the list) and a link to the next node.

struct llist{
    Data d;
    llist *next;
}

I wonder what if I wanted to add my llist to another structure that would represent a list .

struct mainList{
    llist l;
}

I know it may create some difficulties as I'm not quite sure how to pass main list to a function.

Here, I've tried to print linked list

void show(mainlist *ml){
    llist *u = ml->l;
    while(u){
        printf("Date: %s\t Name: %s\n",  u->d.dat, u->d.uname/* u->d.dat, u->d.uname*/);
        u=u->next;
    }
}

But got an error saying that "I cannot 'llist' to 'llist*' in initialization' So, I'm clueless here... Any ideas?

Answer 1

There are many issues - however, one pertaining to the error you are referring is the line:

llist *u = ml->l;  /* I guess you mean struct llist *u = ml->l */

in show function. Here u is a struct llist * , but ml->l is a struct llist , but NOT a pointer to it. You need to change the struct mainList as:

struct mainList{
    struct llist *l;
}

So that ml->l is a struct llist * .

Answer 2

Working solution below, your code snippet had some issues. pointed out in comments...

#include <iostream>
using namespace std;

struct Data {
    int numb;
    int date;
};

struct llist {
    Data d;
    llist *next;
};

struct mainList{
    llist *l; /*should be a pointer as you are referencing it as a pointer*/
};

void show(mainList *ml){ /*should be mainList, your code snippet shows 'mainlist'*/
    llist *u = ml->l;
    while(u){
        printf("Date: %d\t Name: %d\n",  u->d.date, u->d.numb/* u->d.dat, u->d.uname*/); /*your code snippet was using unavailable members of the struct*/
        u=u->next;
    }
}

int main ()
{

    mainList ml;

    show(&ml);

    return 0;
}