Creating a generic linked list in c++ without stl

Question

Can someone help creating generic linkedlist without STL. How do I declare head in main. Is it struct node<>* head ? or struct node* head ? I got a an error using both and it was something like a template declaration cannot appear at block scope

#include <iostream>
using namespace std;


template<class T>
struct node
{
    T data;
    struct node<T>* next;
};

template<class T>
void Push(struct node<T>** H,T dat)
{
    struct node<T> * newnode=(struct node<T> * )malloc(sizeof(struct node<T>)) ;
    newnode->data=dat;
    newnode->next=*H;
    *H=newnode;

}


int main() {

    struct node<>* head=NULL;
    struct node<>* current;
    int a=10;
    float f=10.1;


    Push<int>(&head,a);
    Push<float>(&head,f);


    current=head;
    while(current)
    {
        cout<<current->data;
        current=current->next;
    }

    //code
    return 0;
}

Why are you not using `new` and `delete` instead of `malloc` and `free`? — Rakete1111, Aug 24 '16 at 03:40
node uses stl right? I want to write linkedlist from scratch without using STL much — Niranjan Kotha, Aug 24 '16 at 03:42
Using `malloc` in a simple case like this works, but it only allocate memory. The `new` operator does more (like calling constructors). — Some programmer dude, Aug 24 '16 at 03:42
Also, templates are a *compiler* thing. The standard library uses it extensively, but it's not something library-specific. I think you need to find [a good beginners book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) and start over with that. — Some programmer dude, Aug 24 '16 at 03:43
Finally, `node` is a *template*, it's sort of a blueprint for a class but not a finished class. `node` is a concrete class that can be instantiated into objects. — Some programmer dude, Aug 24 '16 at 03:46
In `main`, you can't use `node<>` because there is no such thing as `node<>`. You could use `node` or `node` depending on whether you want a list of ints or a list of floats. — user253751, Aug 24 '16 at 04:20
I am using template to have a mix of int and float in my linked list. So how do I go about that in main? — Niranjan Kotha, Aug 24 '16 at 04:23
The first thing you need to do for learning C++ is forget most of C. — n. m. could be an AI, Aug 24 '16 at 05:47

score 1 · Accepted Answer · answered Aug 24 '16 at 03:44

First of all, this is a weird mix of C and C++ style programming. But let's ignore that and focus on your real question. Your primary issue is that you're not specifying a type parameter when referencing node (should be node<T> when you use it). So changing that first bit to:

template<class T>
struct node
{
    T data;
    struct node<T>* next; 
};

template<class T>
void Push(struct node<T>** H,T dat) // <-- now we use node<T> everywhere
{
    struct node<T> * newnode=(struct node<T> * )malloc(sizeof(struct node<T>)) ;
    newnode->data=dat;
    newnode->next=*H;
    *H=newnode;

}

Should get you where you need to go. There, you're properly referring to it as node<T> everywhere in Push. Same will apply to main(). Now malloc will work, as a node<T> does have a definite size.

That said, you'll find it a bit cleaner to use node<T> *example = new node<T> and delete example instead.

There are a number of other improvements that could be made to move this more into the C++ realm but I'm just focusing on your direct question here; move on to the rest later.

can you modify the main function too since I am getting some errors like what is the template argument I need to give declaring head. Is it struct node<>* head ? Thanks ! — Niranjan Kotha, Aug 24 '16 at 03:59
@NiranjanKotha The modifications that need to be made to `main()` and everywhere else are exactly the same as the ones shown here. You might want to check out http://www.tutorialspoint.com/cplusplus/cpp_templates.htm. — Jason C, Aug 24 '16 at 05:03

Creating a generic linked list in c++ without stl

1 Answers1