Visual studio 2017 code not outputting after a certain line - c++

Question

I am writing a program that takes 8 user inputted integers and makes a linked list out of them. I have the program print the linked list, then I delete the last node and print the list in reverse. Along the way I've been testing to program to make sure each part works, and it's worked up to the point of printing out the original linked list.
When I finished writing the code for the modification and then printing part, I ran into a problem - the program won't output anything after printing out the original list. So for example, if I input 1,2,3,4,5,6,7,8, it will output:
1
2
3
4
5
6
7
8
and that's it. I've tried putting cout << "testing"; at different points to see where my code stops outputting and the latest point where it successfully outputs is right before the while loop.
I'm not sure why a while loop would cause the program to straight up stop outputting anything, even an arbitrary cout statement that has nothing to do with the while loop itself, so I figured I'd ask on here. I'm using visual studio 2017 if that helps. Thanks for any and all help!
#include <iostream>
using namespace std;
void getdata(int & info); //function that assigns a user inputted value to each node
const int nil = 0;
class node_type // declaration of class
{
public:
int info;
node_type *next;
};
int main()
{
node_type *first, *p, *q, *r, *newnode;
first = new node_type;
newnode = new node_type;
int info;
getdata(info); //first node
(*first).info = info;
(*first).next = nil;
getdata(info); //second node
(*newnode).info = info;
(*first).next = newnode;
(*newnode).next = nil;
p = newnode;
for (int i = 2; i < 8; i++) //nodes 3-8
{
newnode = new node_type;
getdata(info);
(*newnode).info = info;
(*p).next = newnode;
p = newnode;
(*newnode).next = nil;
}
q = first;
while (q != nil) // printing linked list
{
cout << (*q).info << "\n";
q = (*q).next;
}
//deletes last node then reverses list
p = first;
q = (*p).next;
r = (*q).next;
if (first == nil) //if list is empty
cout << "Empty list";
else if ((*first).next == nil) //if list has one node
first = nil;
else if (r == nil) //if list has two nodes
q = nil;
else //general case
{
(*first).next = nil; //last line where when i put a cout << ""; it prints in the output window
while ((*r).next != nil)
{
(*q).next = p;
(*r).next = q;
p = q;
q = r;
r = (*r).next;
}
(*q).next = p;
first = q;
}
q = first;
while (q != nil) // printing newly modified list.
{
cout << (*q).info << "\n";
q = (*q).next;
}
return 0;
}
void getdata(int & info)
{
cout << "Enter number: \n";
cin >> info;
}

Answer 1

else //general case
{
(*first).next = nil; //last line where when i put a cout << ""; it prints in the output window
while ((*r).next != nil)
{
(*q).next = p;
(*r).next = q;
p = q;
q = r;
r = (*r).next;
}
(*q).next = p;
first = q;
}
You run into an infinite loop when reversing your list. That's why it doesn't output anything.
As soon as you link element 2 to element 3, you link element 3 to element 2 in the next step.
That causes an infinite loop when iterating through the list while reversing it.
A proper way to reverse it in the general case would be like that:
else //general case
{
node_type* current = first;
node_type* next = first->next;
(*first).next = nil; //last line where when i put a cout << ""; it prints in the output window
while (next)
{
node_type* temp = next->next;
next->next = current;
current = next;
next = temp;
}
first = current;
}
On a sidenote, you should use bla->foo to dereference, instead of (*bla).foo.