Array to LinkedList in C++

Question

Hi everyone trying to translate an older C program that uses array of structures into C++ program that uses linked lists. I'm a total C++ newb and I'm a little confused on the syntax of setting up a linked list in C++.... here's my code:

#include <iostream> 
#include <stdlib.h>
#include <string>
#include <ctype.h>
#include <fstream>

using namespace std;


struct Video { 
char video_name[1024];      
int ranking;                // Number of viewer hits
char url[1024];             // Video URL
struct Video *next;  // pointer to Video structure
} 


struct Video* Collection = new struct Video;
Collection *head = NULL;                    // EMPTY linked list

In my old program Collection was an array of Video. How can I make Collection be a linkedlist of Video nodes? I am currently getting errors saying on the last two lines code saying: expected initializer before 'Collection' and expected constructor, destructor or type conversion before '*' conversion. I know my syntax is definately wrong, but I guess I dont understand how to create a linked list of Videos inside of Collection...

If you don't know how to implement a linked list, use the STL's List class — Sidharth Mudgal
– Sidharth Mudgal, Commented Oct 7, 2012 at 0:05
Get a book !! Get a book !! Here is some help. stackoverflow.com/questions/388242/… — DumbCoder
– DumbCoder, Commented Oct 7, 2012 at 0:08
@SidharthMudgal Indeed, even if you do know how to implement a linked list, STL's will be solid, tested, and optimized, and with lots of useful helper functions like std::sort already there. Except as a learning exercise, there is zero reason to be writing your own linked list. — Andrew Lazarus
– Andrew Lazarus, Commented Oct 7, 2012 at 0:11

andre · Accepted Answer · 2012-10-07 00:12:38Z

2

The c++ answer is:

struct Video { 
    std::string video_name;     
    int ranking;                // Number of viewer hits
    std::string url;             // Video URL
} 

std::list<Video> list_of_videos

answered Oct 7, 2012 at 0:12

andre

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2 Comments

HXCaine Over a year ago

Agreed -- use STL rather than implementing from scratch

Kevin Anderson Over a year ago

I agree. Here's a link for ease of use: en.cppreference.com/w/cpp/container/list

Marcelo Cantos · Accepted Answer · 2012-10-07 00:16:51Z

0

You've defined Collection as a variable of type pointer-to-Video. On the next line you treat it as a type, which makes no sense. All you need is this:

Video *head = NULL;

head represents the linked list. You don't need another variable.

OTOH, if you really want to use C++ properly, I'd suggested sticking with the array solution unless your usage patterns somehow warrant linked-list semantics. If it's an array of known size, you have two choices:

Video videos[N];
std::array<Video, N> videos; // Preferred in C++11

Otherwise, use a std::vector<T>:

std::vector<Video> videos;

If it really must be a linked list, consider using std::list<T>:

std::list<Video> videos;

In all such cases, you should leave out struct Video *next;.

edited Oct 7, 2012 at 0:16

answered Oct 7, 2012 at 0:11

Marcelo Cantos

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