Doubly Linked List in C, showing head pointer values instead of inserted values-CodePudding

This is the code for a doubly linked list where the values are inserted at the beginning. The code keeps returning the head values instead of the actual values.

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

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

struct Node* head;

void InsertatBegin(int a){
    struct Node* NewNode = (struct Node*)(malloc(sizeof(struct Node*)));
    NewNode->data = a;
    NewNode->next = NULL;
    NewNode->prev = NULL;
    if (head == NULL){
        head = NewNode;
        return;
    }
    NewNode->next = head;
    head->prev = NewNode;
    head = NewNode;

}

void traverse(){
    struct Node* temp = head;
    while (temp != NULL) {
    if (temp->next == NULL) {
        printf(" %d->NULL", temp->data);
        }
        else {
            printf(" %d->", temp->data);
        }
        temp = temp->next; // Traversing the List till end
    }
    printf("\n");
}

int main(){
    head = NULL;
    InsertatBegin(5);
    InsertatBegin(6);
    InsertatBegin(7);
    InsertatBegin(8);
    InsertatBegin(9);
    traverse();

}

This is the output for the code, it seems to print the address of the nodes instead of the values stored in the Node.data structure.

Output:
        752904464-> 752904448-> 752904432-> 752904416-> 5->NULL

CodePudding user response：

struct Node* NewNode = (struct Node*)(malloc(sizeof(struct Node*)));

First step is to remove some excess parentheses!

struct Node* NewNode = (struct Node*) malloc( sizeof struct Node* );

Less code makes it easier to spot what may not be correct... So let's fix that.

struct Node* NewNode = (struct Node*) malloc( sizeof struct Node );

You probably shouldn't be "casting" the return from malloc( ) unless you're using a very old compiler.

struct Node* NewNode = malloc( sizeof struct Node );

struct Node is repeated, and may create a hard-to-find bug if the code is altered carelessly. How much space do we want? Enough to point at with our pointer!

struct Node* NewNode = malloc( sizeof *NewNode );

Easier to read?

And, wars have been fought over this, but this example shows the benefit of shifting the '*' in the declaration of a pointer.

struct Node *NewNode = malloc( sizeof *NewNode );

NewNode is first-and-foremost a pointer. It just happens to be used to point an instance of a "struct Node".

CodePudding user response：

As mentioned in the comments, you should allocate memory as Node not Node*.

struct Node* NewNode = (struct Node*) malloc(sizeof(struct Node));

It allocates enough memory for a Node structure and returns a pointer to that allocated memory. You can take a look at this