Assignment from incompatible pointer type when using Linked Lists

Question

I'm attempting to use a simple linked list example, in attempt to understand the basic idea behind using them, for later use. However, I've gotten confused about how to set each node in the list to a certain value. I.e here, I want to set member "a" to the address of "b" and member "b" to the address of c. However, this is where the warning occurs,

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

struct List
{
    int data;
    struct List * next;
};

int main (int argc, char * argv[])
{

    struct List * a = malloc(sizeof(struct List));
    a->data = 0;
    a->next = NULL;

    struct List * b = malloc(sizeof(struct List));
    b->data = 1;
    b->next = NULL;

    struct List * c = malloc(sizeof(struct List));
    c->data = 2;
    c->next = NULL;

    a->next = &b; //warning occurs here
    b->next = &c;
}

Is there a way to set the values of a->next (a.next) and b->next(b.next) without any warning whatsoever?

You want a->next to point to the space allocated by malloc to which b is pointing. Not to the pointer b. — M.M
– M.M, Commented Mar 17, 2015 at 23:41
So the first one and last one (malloc calls) are fine then? just the * b one? — user3739406
– user3739406, Commented Mar 17, 2015 at 23:44
All the malloc calls are fine, the problems are the two lines giving you the warning — M.M
– M.M, Commented Mar 17, 2015 at 23:48

flogram_dev · Accepted Answer · 2015-03-17 23:27:53Z

1

a->next is of type struct List *.

b is of type struct List *.

&b is of type struct List **.

You probably meant to do this: a->next = b;

answered Mar 17, 2015 at 23:27

flogram_dev

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Steve Fallows · Accepted Answer · 2015-03-17 23:28:53Z

0

b is already a pointer to a struct List so you don't need the & operator.

Just do:

a->next = b;

answered Mar 17, 2015 at 23:28

Steve Fallows

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