Question

I’m still learning C and I must have misunderstood pointers at some point.

I was under the impression that the following lines would copy the memory address stored by l->first to temp. They are both struct list_el* pointers, so I can’t see the problem.

struct list_elt * temp;
temp = l->first;

Running my sample code gives me an infinite loop:

user@machine:~$ gcc question.c 
user@machine:~$ ./a.out | head
append(): l->first->val: 30, l->first->next == NULL: 1
main()  : l->first->val: 30, l->first->next == NULL: 1
print() : l->first->val: 30, l->first->next == NULL: 1
print() : temp->val: 30, temp->next == NULL: 0
print() : temp->val: 30, temp->next == NULL: 0
print() : temp->val: 30, temp->next == NULL: 0
print() : temp->val: 30, temp->next == NULL: 0
print() : temp->val: 30, temp->next == NULL: 0
print() : temp->val: 30, temp->next == NULL: 0
print() : temp->val: 30, temp->next == NULL: 0

Here’s question.c. I’m sorry I couldn’t narrow it down further, every time I do the problem seems to magically go away.

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

struct list_elt {
    int val;
    struct list_elt * next;
};
struct linked_list {
    struct list_elt* first;
};

void print(const struct linked_list * l);
struct linked_list* new_list(void);
void append(struct linked_list* l, int value);


main()
{
    struct linked_list * l;
    l = new_list();
    append(l, 30);
    printf("main()  : l->first->val: %d, l->first->next == NULL: %d\n", l->first->val, l->first->next == NULL);
    print(l);
}

struct linked_list* new_list()
{
    struct linked_list* l;
    l = (struct linked_list*) malloc(sizeof(struct linked_list));
    l->first = NULL;
    return l;
}

void append(struct linked_list* l, int value)
{
    struct list_elt el = {0, NULL};
    el.val = value;
    el.next = NULL;
    if (l->first == NULL) {
        l->first = (struct list_elt*) &el;
        printf("append(): l->first->val: %d, l->first->next == NULL: %d\n", l->first->val, l->first->next == NULL);
    } else {
        printf("append(): Unimplemented\n");
    }
}


void print(const struct linked_list * l)
{
    printf("print() : l->first->val: %d, l->first->next == NULL: %d\n", l->first->val, l->first->next == NULL);
    struct list_elt * temp;
    temp = l->first;
    while (temp != NULL) {
        printf("print() : temp->val: %d, temp->next == NULL: %d\n", temp->val, temp->next == NULL);
        temp = temp->next;
    }
    printf("\n");
}

Thanks.

Answer 1

Your append function is wrong. It is allocating el on the stack, and using the pointer in your list. The problem with this is that the memory is will be overwritten with garbage as soon as another function is called after append, in this case, print. Instead, allocate with malloc, something like:

int append(struct linked_list* l, int value)
{
    struct list_elt *el = malloc(sizeof(struct list_elt));

    if (el)
    {
        el->val = value;
        el->next = NULL;
        if (l->first == NULL) {
            l->first = el;
            printf("append(): l->first->val: %d, l->first->next == NULL: %d\n",
                   l->first->val, l->first->next == NULL);
        } else {
            printf("append(): Unimplemented\n");
        }

        return 0;
    }
    else
    {
        return 1;
    }
}