I was trying to merge two lists and my merge function is not running properly. The program just terminates when I call merge function

Question

I tried putting dummy statements to find out the problem. It seems the problem occurs at the end of the merge() function at p->next=list2; but I do not know what the problem is.

#include <iostream>
using namespace std;

int value, count, value2, count2,choice;

struct node
{
    int data;
    node * next;
};

node *list = nullptr;
node *list2 = nullptr;
node * p;
node * q;
node * r;

void insertFront()
{
    cout << "ENTER A VALUE=";
    cin >> value;

    if (list == nullptr)
    {
        p = new node();
        p->data = value;
        p->next = nullptr;
        list = p;
    }
    else
    {
        p = new node();
        p->data = value;
        p->next = list;
        list = p;
    }
}

void insertFront2()
{
    cout << "ENTER A VALUE=";
    cin >> value;

    if (list2 == nullptr)
    {
        r = new node();
        r->data = value;
        r->next = nullptr;
        list2 = r;
    }
    else
    {
        r = new node();
        r->data = value;
        r->next = list2;
        list2 = r;
    }
}

void delFront()
{
    if (list == nullptr)
    {
        cout << "LIST IS ALREADY EMPTY";
    }
    else
    {
        p = list;
        list = p->next;
        delete(p);
    }
}

void display(int choice)
{
    int select=choice;

    if (select == 1)
    {
        p = list;
    }
    if (select == 2)
    {
        p = list2;
    }
    while (p != nullptr)
    {
        cout << p->data << " ";
        p = p->next;
    }

    cout << endl;
}

void mergeLists(node *list,node *list2)
{
    if (list==nullptr and list2==nullptr){
        cout<<"Both lists are empty";
    }

    else if(list==nullptr && list2!=nullptr){
        display(2);
    }

    else if(list2==nullptr && list!=nullptr){
        display(1);
    }
    if(list!=nullptr && list2!=nullptr){
        p=list;
        while(p!=nullptr){
            p=p->next;
        }

        p->next=list2;

        display(1);
    }
}

int main()
{
    int choice;
    cout << "1) Insert at front " << endl;
    cout << "2) Delete at front" << endl;
    cout << "7) Merge two lists" << endl;
    cout << "9) Display" << endl << endl;
    while (choice != 99)
    {
        cout << "Your choice:";
        cin >> choice;

        switch (choice)
        {
            case 1:
                {
                    int sel;
                    cout << "Enter the list to which you want to enter:\n 1 or 2\nYour choice:";
                    cin >> sel;
                    if (sel == 1)
                    {
                        insertFront();
                    }
                    else if (sel == 2)
                    {
                        insertFront2();
                    }

                    break;
                }

            case 2:
                {
                    delFront();
                    break;
                }

            case 7:
                {
                    mergeLists(list,list2);

                    break;
                }

            case 9:
                {
                    int sel;
                    cout<<"Select a list to display: 1/2"<<endl;
                    cin>>sel;
                    display(sel);
                    break;
                }

            case 99:
                {
                    cout << "PROGRAM TERMINATED :)";
                    break;
                }
        }
    }

    return 0;
}

Answer 1

Inside merge(), you are looping through list until p becomes nullptr, thus p is nullptr when you try to access p->next after the loop.

You need to fix your loop to stop when it reaches the last node, not when it goes past the last node, eg:

void mergeLists(node *list, node *list2)
{
    if (!list && !list2){
        cout << "Both lists are empty";
    }

    else if (!list && list2){
        display(2);
    }

    else if (!list2 && list){
        display(1);
    }

    else {
        p = list;
        while (p->next){
            p = p->next;
        }

        p->next = list2;

        display(1);
    }
}

Answer 2

For starters if you want to append the second list list2 to the first list list then after the calling the function merge the list list2 should be empty.

Within the function merge after this while loop

    while(p!=nullptr){
        p=p->next;
    }

the pointer p is a null pointer. So using the null pointer to access memory in this statement

    p->next=list2;

results in undefined behavior.

And this if statement

if(list!=nullptr && list2!=nullptr){

is redundant.

Using your approach the function can be defined the following way

void mergeLists(node *list,node *list2)
{
    if ( list == nullptr and list2 == nullptr )
    {
        cout<<"Both lists are empty";
    }
    else if ( list == nullptr && list2 != nullptr )
    {
        
        list = list2;
        list2 = nullptr; 
        display(1);
    }
    else if ( list2 == nullptr && list != nullptr )
    {
        display(1);
    }
    else
    {
        p = list;

        while ( p->next != nullptr )
        {
            p = p->next;
        }

        p->next = list2;
        list2 = nullptr;

        display(1);
    }
}

Pay attention to that it is a bad idea to make functions depend on global variables. For example if you will want to use one more list you will need to rewrite your functions.

Also you have too much redundant code as for example in the functions insertFront and insertFront2. They can be defined much simpler. For example

void insertFront()
{
    cout << "ENTER A VALUE = ";
    cin >> value;

    list = new node { value, list };
} 

and

void insertFront2()
{
    cout << "ENTER A VALUE = ";
    cin >> value;

    list2 = new node { value, list2 };
}