To delete the first node of a doubly linked list using C, you can follow these steps:

1. Check if the list is empty. If it is, there is nothing to delete, so return from the function.

```
if (*headRef == NULL) {
    printf("List is empty.\n");
    return;
}
```

Here, `headRef` is a pointer to the pointer to the first node of the list.

2. If the list is not empty, store the address of the first node in a temporary variable and update the head pointer to point to the second node.

```
struct Node* temp = *headRef;
*headRef = temp->next;
```

3. If the second node exists, update its previous pointer to `NULL` because it is now the first node.

```
if (*headRef != NULL) {
    (*headRef)->prev = NULL;
}
```

4. Free the memory occupied by the first node using the `free()` function.

```
free(temp);
```

The complete code for deleting the first node of a doubly linked list using C is shown below:

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

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

void deleteFirstNode(struct Node** headRef) {
    if (*headRef == NULL) {
        printf("List is empty.\n");
        return;
    }

    struct Node* temp = *headRef;
    *headRef = temp->next;

    if (*headRef != NULL) {
        (*headRef)->prev = NULL;
    }

    free(temp);
}

void printList(struct Node* node) {
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}

int main() {
    struct Node* head = NULL;
    struct Node* node1 = (struct Node*)malloc(sizeof(struct Node));
    struct Node* node2 = (struct Node*)malloc(sizeof(struct Node));
    struct Node* node3 = (struct Node*)malloc(sizeof(struct Node));

    node1->data = 1;
    node1->prev = NULL;
    node1->next = node2;

    node2->data = 2;
    node2->prev = node1;
    node2->next = node3;

    node3->data = 3;
    node3->prev = node2;
    node3->next = NULL;

    head = node1;

    deleteFirstNode(&head);

    printList(head); // Output should be: 2 3

    return 0;
}
```

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