CODEWITHSUNDEEP
cfunctionpointers
shost71
shost71

Reputation: 85

change a pointer that I passed inside a function in C

I have this piece of code which will (hopefully) become a singly-linked list implementation.

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

struct NODE{
   int d;
   struct NODE *next;
};

int addNode(int n, struct NODE **root);

int main(){

    struct NODE *root = NULL;
    addNode(3, &root);

    printf("%i\n", root->d);    

    getch();
    return 0;
}

int addNode(int n, struct NODE **root){


   if(*root == NULL){
           *root = malloc(sizeof(struct NODE));
           *root->d = n;
           *root->next = NULL;
           return 0;
   }
}

When I run it, I get "request for member 'd' in something not a structure or union" inside the addNode function; same with the 'next' part. If instead I change the function to this:

int addNode(int n, struct NODE **root){

struct NODE *temp;

if(*root == NULL){
        *root = malloc(sizeof(struct NODE));
        temp = *root;
        temp->d = n;
        temp->next = NULL;
        return 0;
    }
}

it works just fine. My question is; why do I have to create temp? As I understand it, in the first version of the code I am passing a pointer to root, so that there should be no 'pass by reference' issues, and it should not have problems running... What's the mistake?

Upvotes: 0

Views: 130

Answers (3)

Jay
Jay

Reputation: 24895

Please refer to the Operator Precedence Table. The -> operator has a higher precedence than the * (dereference) operator, which is the reason for your error.

*root->d is interpreted as *(root->d) instead of (*root)->d due to operator precedence.

Upvotes: 0

Raymond Chen
Raymond Chen

Reputation: 45173

Operator precedence.

*root->d = n;

is parsed as

*(root->d) = n;

But you want

(*root)->d = n;

Upvotes: 1

Karthik T
Karthik T

Reputation: 31952

(*root)->d = n;
(*root)->next = NULL;

Its an operator precedence problem. Using () will fix it.

Upvotes: 3

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