为什么我们使用指向一个的指针和指向另一个的普通指针？

Question

// A complete working C++ program to demonstrate
// all insertion methods on Linked List
#include <bits/stdc++.h>
using namespace std;

// A linked list node
class Node
{
    public:
    int data;
    Node *next;
};

/* Given a reference (pointer to pointer)
to the head of a list and an int, inserts
a new node on the front of the list. */
void push(Node** head_ref, int new_data)
{
    /* 1. allocate node */
    Node* new_node = new Node();

    /* 2. put in the data */
    new_node->data = new_data;

    /* 3. Make next of new node as head */
    new_node->next = (*head_ref);

    /* 4. move the head to point to the new node */
    (*head_ref) = new_node;
}

/* Given a node prev_node, insert a new node after the given
prev_node */
void insertAfter(Node* prev_node, int new_data)
{
    /*1. check if the given prev_node is NULL */
    if (prev_node == NULL)
    {
        cout<<"The given previous node cannot be NULL";
        return;
    }

    /* 2. allocate new node */
    Node* new_node = new Node();

    /* 3. put in the data */
    new_node->data = new_data;

    /* 4. Make next of new node as next of prev_node */
    new_node->next = prev_node->next;

    /* 5. move the next of prev_node as new_node */
    prev_node->next = new_node;
}

/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
void append(Node** head_ref, int new_data)
{
    /* 1. allocate node */
    Node* new_node = new Node();

    Node *last = *head_ref; /* used in step 5*/

    /* 2. put in the data */
    new_node->data = new_data;

    /* 3. This new node is going to be
    the last node, so make next of
    it as NULL*/
    new_node->next = NULL;

    /* 4. If the Linked List is empty,
    then make the new node as head */
    if (*head_ref == NULL)
    {
        *head_ref = new_node;
        return;
    }

    /* 5. Else traverse till the last node */
    while (last->next != NULL)
    {
        last = last->next;
    }

    /* 6. Change the next of last node */
    last->next = new_node;
    return;
}

// This function prints contents of
// linked list starting from head
void printList(Node *node)
{
    while (node != NULL)
    {
        cout<<" "<<node->data;
        node = node->next;
    }
}

/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
    
    // Insert 6. So linked list becomes 6->NULL
    append(&head, 6);
    
    // Insert 7 at the beginning.
    // So linked list becomes 7->6->NULL
    push(&head, 7);
    
    // Insert 1 at the beginning.
    // So linked list becomes 1->7->6->NULL
    push(&head, 1);
    
    // Insert 4 at the end. So
    // linked list becomes 1->7->6->4->NULL
    append(&head, 4);
    
    // Insert 8, after 7. So linked
    // list becomes 1->7->8->6->4->NULL
    insertAfter(head->next, 8);
    
    cout<<"Created Linked list is: ";
    printList(head);
    
    return 0;
}


// from geeksforgeek

Why the append and push function have a pointer to pointer (head_ref), but the insertAfter function has a normal pointer (prev_node)? As we are changing them whether it is the head or the previous node, so it should be same approach(types of pointer).

Can I do use a normal pointer instead of a pointer to pointer for the same?

Answer 1

The functions push and append both need to be able to change the pointer to the head node.

If they were only passed a Node* pointing to the head, then they would only be able to change their local copy of that pointer to the head node. However, they need to be able to change the original pointer to the head node, which is the variable head in the function main . That is why they need a pointer to the variable head in the function main , so both functions need to be passed a Node** .

In contrast to the functions push and append , the function insertAfter will never need to change the pointer that is being passed to the function. That is why it is sufficient to pass a Node* .