[C]: When a pointer to a pointer to a struct is passed in a function, why can't it change the elements of that struct?

Question

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

void insertAfterHead(struct Node **a, int x) {
struct Node *newNode = (struct Node *)malloc(sizeof(struct Node));
newNode -> data = x;
newNode -> next = *a -> next;
*a -> next = newNode;
}


int main() {
    struct Node *head = NULL;
    struct Node *second = NULL;
    struct Node *third = NULL;

    head = malloc(sizeof(struct Node));
    second = malloc(sizeof(struct Node));
    third = malloc(sizeof(struct Node));

    head -> data = 5;
    head -> next = second;

    second -> data = 10;
    second -> next = third;

    third -> data = 15;
    third -> next = NULL;

    insertAfterHead(&head, 10);

    return 0;
}

This is the error message I get on running the code:

error: member reference base type 'struct Node *' is not a structure or union newNode -> next = *a -> next;

error: member reference base type 'struct Node *' is not a structure or union *a -> next = newNode;

Please be as descriptive as possible as I'm a beginner to C :)

The error message is not corresponding to the code. And paste it here, not as an image but as a text. — Eugene Sh., Dec 13 '17 at 18:55
Please post your output (error message) as text, and please post the error message you got from the code you posted (not from some other code). — ikegami, Dec 13 '17 at 18:56
@EugeneSh. Sorry, I renamed the variables (for clarification) while asking. Added error message too. — Apoorve, Dec 13 '17 at 19:00
@JohnnyMopp - Yes but I forgot the context in which I wanted to ask this so I made it up :P — Apoorve, Dec 13 '17 at 19:03

score 2 · Accepted Answer · answered Dec 13 '17 at 18:59

2

*headRef->next

means

*(headRef->next)

but you want

(*headRef)->next

answered Dec 13 '17 at 18:59

ikegami

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[C]: When a pointer to a pointer to a struct is passed in a function, why can't it change the elements of that struct?

1 Answers1