如何在不使用 C++ 函数的情况下显示链表中的元素？

Question

I want to display elements in the linked list without using function in C++. My code does not appear anything on the screen, even it does not contains any errors. I do not know how can I fix it?

#include <iostream>
using namespace std;

struct Element {
    int data;
    Element *next;
};

struct List {
    int nb_ele;
    Element *head;
    Element *tail;
};

int main(){
    Element *tmp;

    tmp = new(Element);
    tmp ->data = 5;
    tmp ->next = NULL;

    List *li;
    li ->head = tmp;
    li ->tail = tmp;
    li ->nb_ele = 1;

    tmp = new(Element);
    tmp ->data = 7;
    tmp ->next = li->head;
    li ->head = tmp;
    li ->nb_ele = li->nb_ele + 1;

    Element *ptr;
    ptr = li->head;
    while (ptr != NULL){
        cout<< ptr->data<<" ";
        ptr = ptr->next;
    }
}

Answer 1

The problem with your code is this line:

List *li;

li is an uninitialized pointer and when you try to access it later, causes problems. change this line to:

List *li = new List;

or don't use a pointer. simply change it to:

List li;