How to implement linked list in reference without using pointer

Question

In Java, a reference can be initialized to null. However, in C++, it will cause a problem. So not sure how to use reference-only to implement the linked list.

@lightrek Have you considered using smart pointers. Like unique_ptr? — bashrc
– bashrc, Commented May 17, 2016 at 5:09
Possible duplicate of Is a Linked-List implementation without using pointers possible or not? — ibezito
– ibezito, Commented May 17, 2016 at 5:10
A Java reference is much more similar to a C++ pointer than to a C++ reference. They're different languages, they use different terminology. — Benjamin Lindley
– Benjamin Lindley, Commented May 17, 2016 at 5:16
References in Java are like pointers in C++ so use pointers. — Galik
– Galik, Commented May 17, 2016 at 5:17

Eissa N. · Accepted Answer · 2016-05-17 06:18:53Z

2

I don't know how useful the concept is, but you CAN do it using std::reference_wrapper, as follows:

#include <iostream>
#include <list>
#include <functional>
using namespace std;

int main() {
    int a = 2, b = 6, c = 1;

    list<reference_wrapper<int>> mylist;
    mylist.push_back(a);
    mylist.push_back(b);
    mylist.push_back(c);

    for(auto x : mylist) {
        cout << x << " ";
    }
    cout << endl;
    a = 3; // <- this setting will modify mylist!

    for(auto x : mylist) {
        cout << x << " ";
    }
    return 0;
}

I would recommend learning C++ ways of handling things, specially that you are coming from Java world. Demo!

edited May 17, 2016 at 6:18

answered May 17, 2016 at 5:24

Eissa N.

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2 Comments

Ajay Over a year ago

While your answer is correct. It may be beyond the understanding of OP, who isn't well aware of pointers/references in C++. No offense to anyone!

Eissa N. Over a year ago

@Ajay I agree. But, I just wanted to show possibilities :)

score 2 · Accepted Answer · 2016-05-17 07:51:00Z

2

Im not sure why you would want to use references in c++ since a reference cannot be null in C++ like you said. What would you do if you get to the end of the linked list.

Your only solution (Since you are new to c++) is to use pointers like this.

struct Node{
   int value;
   Node* next;
}

This way you can leave the pointer called next as null and that would signify the end of the linked list.

edited May 17, 2016 at 7:51

answered May 17, 2016 at 5:10

user4603522

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