Linked lists

[Dhanish]

I need to write a program to concatenate two linked lists.
I have to include functions to:
- create the two lists
- display each list
- concatenate the two lists
- display the new list created after the concatenation process.

Here is what i managed to do till now

#include <cstdlib>
#include <iostream>

using namespace std;

struct link
    {
           char *data;
           link *next;
    };
    
    typedef struct link node;
    node *head;

struct link2
    {
           char *data2;
           link2 *next;
    };
           
    typedef struct link2 node2;
    node *head2, *current, *templist;
    
    char a[30];

    void populatelist();
    void displaylist();

int main()
{    
    head = new node;
    head->data = "Name: ";
    head->next = NULL;
    
    head->next = new node;
    head->next->data = "Surname: ";
    head->next->next = NULL;
    
    head->next->next = new node;
    head->next->next->data = "Telephone: ";
    head->next->next->next = NULL;
    
    cout << "List 1 has been created\n\n";
    
    cout << head->data <<endl;
    cout << head->next->data <<endl;
    cout << head->next->next->data <<endl;
    cout << "\n\n";
    
    head2 = NULL;
    
    populatelist();
    
    cout << "Data Entered\n\n";
    displaylist();

    system("pause");
    return EXIT_SUCCESS;
}

void populatelist()
{
     cout << "Enter Name, Surname and then Address\n\n";
     for(int i=0;i<3;i++)
     {
           cout << "->Enter data: ";
           cin >> a;
           current = new node;
           current->data2=a;
           current->next = head2;
           head2 = current;
           cout << "Data has been populated...next data\n\n";
     }
}

void displaylist()
{
     templist = head2;
     while(templist != NULL)
     {
          cout << "-->" << templist->data <<endl;
          templist=templist->next;
     }
     
     cout << "\n";
}

PLZ HELP!

[1yR4jE8b]

That's easy, just change the link pointer of the last element of the first list, to the first element of the second list, done.

[Dhanish]

im new at programming and i barely managed to do this much by going over several tutorials
i don't think my 2nd list is even being read
because as output i only get the phone number displayed thrice

[boolivar]

Mistake here(line 69):

current->data2=a;

All nodes in 2nd list take pointer to the same array a[].

Solution may be like this:

current->data2 = new char [strlen(a)+1];  //plus 1 for null-character
strcpy(current->data2, a);

*don't forget release memory afterwards,
may be like this:

void DisposeList()
{
node2 *templist;
node2 *templist_next;

	templist = head2;

	while (templist)
	{
		templist_next = templist->next;
		delete [] templist->data2;
		delete templist;
		templist = templist_next;
	}
}

** here a misprint(line 22) ?

node *head2, *current, *templist;

mean:

node2 *head2, *current, *templist;