C linked list incompatible pointer type

Question

struct list_node {
    int value;
    struct list_node *next;
};

struct linked_list {
    int size;
    struct list_node *head;
};

void print_linked_list(struct linked_list *list){
    struct linked_list *current = list;

    while(current != NULL){
        printf("%d ", current->head->value);
        current = current->head->next;
    }
}

I have to define a function to print out a linked list, but I get an error message saying "incompatible pointer type". I know that the problem is in "current = current->head->next;" but how can I achieve that?

You need struct list_node *curr = list->head; in place of struct linked_list *current = list;, and consequential changes in the loop (while (curr != NULL) { printf("%d ", curr->value); curr= curr->next; }). — Jonathan Leffler
– Jonathan Leffler, Commented Jan 17, 2022 at 14:25

ryyker · Accepted Answer · 2022-01-17 14:28:38Z

3

current is a struct linked_list*, but current->head->next is a struct list_node*.

struct linked_list and struct list_node are two different unrelatd structures in spite of their similarity.

You cannot assign pointers to different types, hence the error message incompatible pointer type.

You probably want this:

void print_linked_list(struct linked_list* list) {
  struct list_node* current = list->head;

  while (current != NULL) {
    printf("%d ", current->value);
    current = current->next;
  }
}

edited Jan 17, 2022 at 14:28

ryyker

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answered Jan 17, 2022 at 14:24

Jabberwocky

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Comments

Vlad from Moscow · Accepted Answer · 2022-01-17 14:31:58Z

1

The function

void print_linked_list(struct linked_list *list){
    struct linked_list *current = list;

    while(current != NULL){
        printf("%d ", current->head->value);
        current = current->head->next;
    }
}

does not make a sense.

In this statement

current = current->head->next;

the pointer current has the type struct linked_list * while the expression current->head->next has the type struct list_node *.

It seems you mean

void print_linked_list( const struct linked_list *list )
{
    if ( list != NULL )
    {
        for ( const struct list_node *current = list->head; current != NULL; current = current->next )
        {
            printf( "%d ", current->value );
        }
    }
}

edited Jan 17, 2022 at 14:31

answered Jan 17, 2022 at 14:26

Vlad from Moscow

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