Question

我编写了一个代码，用于创建由用户指定并打印的大小的链接列表。但我注意到了一些奇怪的行为：

如果我实现这个算法，它会正确打印一个列表：

struct num

{ int l;
num* p_next; };

void add (int size, num* head) {

    srand (time(0));
    for (int i = 0; i< size; i++) {
    num* newnode = new num;
    newnode->l = rand()%100;
    newnode->p_next = head;
    head = newnode;}

while (head != 0) {
    cout << head->l << endl;
    head = head -> p_next;}
    }


int main () {
    num* head = 0;
    add (10, head);
}

但是，当我想通过消除内存泄漏并编写相同的任务时，使代码更安全：

struct num
{ int l;
num* p_next; };

void add (int size, num* head) {
    srand (time(0));
    for (int i = 0; i< size; i++) {
    num* newnode = new num;
    newnode->l = rand()%100;
    newnode->p_next = head;
    head = newnode;}
    }

int extract (num*head) {
    int n;
    n = head->l;
    head = head->p_next;
    delete head;
    return n;}

void print (num *head) { 
while (head != 0) {
        cout << extract (head) << endl;
}}

int main () {
    num* head = 0;
    add (10, head);
    print (head);
cin.get();
cin.ignore();
}

..它不会打印任何东西。我觉得问题在于非常小的问题，但我找不到它。为什么不在第二种情况下打印它？

Answer 1

您的问题是head按值传递（是的，指针可以按值传递）到add而不是通过引用传递。当您致电print时，您的head不在列表的前面（因为它在add中没有实际修改，只有它的副本。）

试试这个：

void add (int size, num*& head) 
{
   srand (time(0)); //actually move this line to main, don't do it here
   for (int i = 0; i< size; i++) 
   {
      num* newnode = new num;
      newnode->l = rand()%100;
      newnode->p_next = head;
      head = newnode;
  }
}

extract和print

存在类似问题

编辑：您的提取功能也会不正确地删除内存。这是一个完整的工作示例：

#include <iostream>
#include <ctime>
using namespace std;

struct num
{
   int l=0;
   num* p_next=nullptr;
};

void add (int size, num*& head) 
{
    for (int i = 0; i<size; i++) 
    {
       num* newnode = new num;
       newnode->l = rand()%100;
       newnode->p_next = head;
       head = newnode;
    }
}

int extract (num*& head)
{
    int n;
    n = head->l;
    num* tmp = head;
    head = head->p_next;
    delete tmp;
    return n;
}

void print (num*& head) 
{ 
   while (head != nullptr)
        cout << extract(head) << endl;
}

int main () 
{
    srand (time(0));
    num* head = nullptr;
    add (10, head);
    print (head);
}

Live Demo

Answer 2

您需要一个临时节点。您需要将temp设置为从头开始的下一个节点，然后删除head。然后将头设置为临时节点。

int extract (num*head) 
{
    int n;
    n = head->l;
    num* temp = head->p_next
    delete head;
    head = temp;
    return n;
}

链接列表未打印

2 个答案: