Allocating space to a node pointer

Question

I am currently attempting to use a doubly linked list to sort some data. I am having trouble creating a new node with the given data. Below was the code given to me:

#ifndef LIST_H_
#define List_H_
#define MAX_SYMBOL_LENGTH 7
struct order {
    int id;
    char symbol[MAX_SYMBOL_LENGTH];
    char side;
    int quantity;
    double price;
};

typedef struct order* OrderPtr;
typedef struct onode* NodePtr;

struct onode {
    OrderPtr data;
    NodePtr next;
    NodePtr prev;
};

This is the code that I have written using list.h as a header. Here is the code that seemingly keeps crashing:

#include "list.h"

NodePtr newNode(OrderPtr data){

    NodePtr node = (NodePtr)malloc(sizeof(NodePtr));
    //node->data = (NodePtr)malloc(sizeof(OrderPtr));
    //*node->data = *data;
    node->data = data;//This is the one I am having problems with
    node->next = NULL;
    node->prev = NULL;
    return node;
}

It compiles fine but when I try and submit it to an online grader it says that it does not work. Here is my thought process,

1. create memory for NodePtr.
2. create memory for NodePtr->data.

and then assign the values of data passed from the function to the values in Node->Ptr. But I do not know how to allocate memory for NodePtr->data.

Answer 1

NodePtr node = (NodePtr)malloc(sizeof(NodePtr));

Isn't doing what you are thinking. It's allocate space to hold a pointer same as sizeof(int*), it's 4 bytes on 32-bit machine, usually.

You need to do NodePtr node = malloc(sizeof(struct onode)); instead of. data member should be result to a malloc(sizeof(struct order));

Also, don't cast result value from a malloc() call.

Answer 2

NodePtr is a pointer to a node and not the node itself. You're only allocating enough memory for a pointer and not all the members of the onode structure. You'll want to call malloc with sizeof(struct onode).