Question

I have a function declared something like

void func1(int& x) {
    func2(x); // func2 accepts an int
}

I think this is what crashes the program? I get the error

R6010 - abort() has been called

What do I need to do to pass x into a function that accepts an int? I expected them to work the same … since I can just echo the value of x using cout << x

UPDATE

Just a test:

cout << stmtNo << endl;
Node* n = ast->getNode(stmtNo);
cout << n->getNodeType() << " " << n->getStmtNo() << endl;

Above fails … Below passes

cout << stmtNo << endl;
Node* n = ast->getNode(1);
cout << n->getNodeType() << " " << n->getStmtNo() << endl;

Answer 1

There is no problem, you can always pass an integer reference as an argument for an integer.

Integer references can be interpreted as constant pointers who automatically de-reference themselves.

#include<iostream>
using namespace std;
void fun_2(int s)
{
   cout<<s<<endl;
}
void func(int &d)
{
    fun_2(d);
}

int main()
{
   int x=99;
   func(x);

   system("pause");
   return 0;
}

The above code works perfectly!