How to pass a variable to a function without using global variables?

Question

C++ / C#

The openGL(glut) packages have a function of the form:
void glutDisplayFunc( void (*func)( ) )
That is, the function accepts a function of type void with no arguments.
And here's the question.
How to pass a variable to a function (which we pass to glutDisplayFunc) without using global variables?
For example, I wrote a simple program illustrating the problem.

#include <glut.h> //Подключаем библиотеку glut
int WIDTH = 400; //Глобальная переменная ширины
int HEIGHT = 400; //Глобальная переменная высоты
void setPixel(int x, int y) //Функция отрисовки пикселя с координатами (x,y)
{
  glBegin(GL_POINTS);
  glVertex2i(x, y);
  glEnd();
}
void myDisplay() //Функция, которая будет отвечать за рисование в окне
{
  glClear(GL_COLOR_BUFFER_BIT);
  glColor3f(0, 0, 0);
  glPointSize(100);
  setPixel(WIDTH / 2, HEIGHT / 2);//Ставим пиксель в центре окна с использованием глобальных переменных
  glFlush(); 
}
void main()
{
  glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB);
  glutInitWindowSize(WIDTH, HEIGHT);//Инциализируем размер окна с использованием глобальных переменных
  glutCreateWindow("Problem");
  glClearColor(1, 1, 1, 1);
  glMatrixMode(GL_PROJECTION);
  gluOrtho2D(0, WIDTH, 0, HEIGHT);//Устанавливаем двумерную область просмотра с использованием глобальных переменных
  glutDisplayFunc(myDisplay); //Собсвенно вот она - проблема
  glutMainLoop();
}

here you can see how to install glut

Answer 1

assigning a function to a variable, then passing it

void (*fcnPtr)(void) = boo;
glutDisplayFunc( fcnPtr );

Answer 2

Read about functional classes. Those. those with operator() overloaded. That is, you set your parameters in the constructor, and then pass the instance itself, which can be called as a function, where it is necessary.
PS See my examples in the last comment...

Answer 3

This function definitely has a friend, which is called similarly, but accepts void*.
Exactly the same void* is received by the function that registers this callback, after which this void*, in fact, gets into the callback.