为什么我们对一个指针使用指针，而对另一个指针使用普通指针？

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

为什么append和push函数有一个指向指针的指针(head_ref)，而insertAfter函数有一个普通的指针(prev_node )?由于我们正在改变它们，无论是头部还是前一个节点，所以它应该是相同的方法（指针类型）。

我可以使用普通指针代替指向指针的指针吗？

Answer 1

函数push和append都需要能够改变指向头节点的指针。

如果只向它们传递一个指向头的 Node*，那么它们只能将该指针的本地副本更改为头节点。但是，他们需要能够改变原来指向头节点的指针，也就是函数main中的变量head。这就是为什么他们需要一个指向函数 main 中的变量 head 的指针，所以两个函数都需要传递一个 Node**.

与函数 push 和 append 相比，函数 insertAfter 永远不需要更改传递给函数的指针。这就是为什么传递 Node*.

就足够了

为什么我们对一个指针使用指针，而对另一个指针使用普通指针？

Why are we using pointer to pointer for one and a normal pointer to another?

c++

pointers

linked-list

data-structures