Returning a pointer from a function using template in c++

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Line inside main Node<int> *head = takeInput() is giving error no matching function for call to 'takeInput()'. Can you help me and tell me whats the mistake I am doing here so that I wouldn't do it in future. Thanks!

#include<iostream>
using namespace std;

template<class T>
class Node{

public: 
    T data;
    Node *next;
    Node(T x){
        data = x;
        next = NULL;
        }
    
};

template<typename T>
Node<T> * takeInput(){
    T data;
    cin>>data;
    Node<T> *head = NULL;
    Node<T> *tail = NULL;
    
    while(data != -1){
        Node<T> *newNode = new Node(data);
        
        if(head == NULL){
            head = newNode;
            tail = newNode;
        }else{
            tail->next = newNode;
            tail = tail->next;
            }
        cin>>data;
        }
    return head;
    }
    
int main(){
    Node<int> *head = takeInput();
    cout<<head->data;
    }
program-entry-point pointers linked-list function c++
1 Answers

Your code has flaws;

Your compare head == NULL, it's always be null, every time function runs because, you assigning NULL Node<T> *head = NULL; at start of function.

And using NULL not recommended in c++, use nullptr;

Took liberty and modified your function. there is many ways you can implement below function, for now i took it below way, I am not saying it's the best way.

template<typename T>
void takeInput( Node<T>*& head, T val ){

    Node<T>* newNode = new Node<T>( val );

    if ( head == nullptr )
    {
        head = newNode;
    }
    else
    {
        Node<T>* tmp = head;

        while ( tmp->next != nullptr ) tmp = tmp->next;

        tmp->next = newNode;

    }
}
    
int main()
{
    Node<int> *head;
    
    takeInput( head, 10 );
    takeInput( head, 20 );
    takeInput( head, 30 );

    std::cout<< head->data << std::endl;
    std::cout<< head->next->data << std::endl;
    std::cout<< head->next->next->data << std::endl;

}
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