What is the difference between Node *head versus Node **head?

问题

I am writing a C code to find the middle of linked list. I understood the logic but was unable to figure out how pointers are being used. What is the difference in the way Node *head and Node** head_ref work?

void middle(struct Node *head) ;

void push(struct Node** head_ref, int new_data) ; 

回答1:

In the first function header, *head is a pointer to a node object which is allocated somewhere in memory:

void middle(struct Node *head);
              _____________________
             |                     |
   *head --> |     Node object     |
             | [val=1][*next=NULL] |
             |_____________________|

while in the second function header, **head_ref is a pointer to a pointer to a node object somewhere in memory:

void push(struct Node** head_ref, int new_data); 
              _____________________
             |                     |
   *head --> |     Node object     |
     ^       | [val=1][*next=NULL] |
     |       |_____________________|
     |    
 **head_ref

This is another layer of indirection. Why is the second construct necessary? Well, if I want to modify something allocated outside of my function scope, I need a pointer to its memory location. In the first example, I can dereference the *head pointer (with head->) to access the underlying Node object in memory and make modifications to it or access its properties.

Taking this logic to the next level of indirection, if I want to push a new Node object onto the front of the list to make it the new head, I need to modify the head pointer itself. Just as when I wanted to manipulate the Node object with a pointer, now I need a pointer to *head (which simply happens to be a pointer rather than a Node object) in order to modify it.

Here's the contents of push and a before/after function call:

void push(struct Node** head_ref, int new_data) {
    Node *new_head = malloc(sizeof(Node)); // allocate memory for the new head node
    new_head->data = new_data;             // set its value

    new_head->next = *head_ref;            // make it point to the 
                                           // old head pointer

    *head_ref = new_head;                  // make it the new head by modifying
                                           // the old head pointer directly
}

/* Before push */
              _____________________
             |                     |
   *head --> |     Node object     |
     ^       | [val=1][*next=NULL] |
     |       |_____________________|
     |    
 **head_ref

/* After calling push(&head, 2);
 *                    ^
 *                   `&` operator gets the memory address of `head`, 
 *                       which is a pointer.
 */

              _________________      _____________________
             |                 |    |                     |
   *head --> |   Node object   |    |      Node object    |
     ^       | [val=2] [*next]----->| [val=1][*next=NULL] |
     |       |_________________|    |_____________________|
     |    
 **head_ref

Here's a full example:

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

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

void push(Node** head_ref, int new_data) {
    Node *new_head = malloc(sizeof(Node));
    new_head->data = new_data;
    new_head->next = *head_ref;
    *head_ref = new_head; 
}

void print(Node *head) {
    while (head) {
        printf("%d->", head->data);
        head = head->next;
    }

    puts("NULL");
}

void free_list(Node *head) {
    while (head) {
        Node *tmp = head;
        head = head->next;
        free(tmp);
    }
}

int main() {
    Node *head = malloc(sizeof(Node));
    head->next = NULL;
    head->data = 1;
    printf("Before push:\n");
    print(head);
    push(&head, 2);
    printf("\nAfter push:\n");
    print(head);
    free_list(head);
    return 0;
}

Output:

Before push:
1->NULL

After push:
2->1->NULL

回答2:

struct Node *head - > Here the pointer head is a pointer to the head of the Linked List. Head is a pointer which can point to a node structure.

eg.

If you have a linked list like this :- ____ ____ _____ |_1__|--->|_2__|--->|_3__|--->....... address- 1000 1004 1008

Your Node *head will be a pointer variable holding the address of the node with value 1(address of the head node which is the node storing the value 1). Content of head=1000

struct Node **head_ref -> here the head_ref is the pointer to the pointer of the start of the linked list.

eg.

If you have a linked list like this :- ____ ____ _____ |_1__|--->|_2__|--->|_3__|--->....... address- 1000 1004 1008 | *head address- 5050 | **head

Your Node *head will be a pointer variable holding the address of the node with value 1(address of the head node which is the node storing the value 1). Content of head=1000 Content of **head_ref will be the address of the head pointer i.e, 5050.

**head_ref is used for indirect reference.

来源：https://stackoverflow.com/questions/55654934/what-is-the-difference-between-node-head-versus-node-head