為什么我們使用指向一個的指針和指向另一個的普通指針？

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

為什么 append 和 push 函數有一個指向指針的指針（head_ref），而 insertAfter 函數有一個普通的指針（prev_node）？ 由於我們正在更改它們是頭節點還是前一個節點，因此應該是相同的方法（指針類型）。

我可以使用普通指針而不是指向指針的指針嗎？

Answer 1

函數push和append都需要能夠改變指向頭節點的指針。

如果他們只傳遞了一個指向頭Node* ，那么他們將只能將該指針的本地副本更改為頭節點。 但是，它們需要能夠更改指向頭節點的原始指針，即函數main的變量head 。 這就是為什么他們需要一個指向函數main變量head的指針，所以兩個函數都需要傳遞一個Node** 。

與push和append函數insertAfter ，函數insertAfter永遠不需要更改傳遞給函數的指針。 這就是為什么傳遞一個Node*就足夠了。