Explanation of typecasting of pointer in C++

I do not understand the two lines mentioned below in the code that I have provided here. Why I need to use int*? How I am accessing the private variable? I am not sure what I am doing with these two lines. Please explain in detail.

Problematic lines:

int *p = (int *) &s;  
*p=10;

Main Code:

#include <iostream>

using namespace std;

class sample {
private:
  int a;
public:
  void function () {
    a=5;
  }
  void printA () {
    cout<<"value is "<<a<<endl;
  }
};

int main () {
  sample s;
  s.function();
  s.printA();

  int *p = (int *) &s;
  *p=10;

  s.printA();
}

Solution

Using (int*) there is a really really bad idea. You are using a class, that isn't an int, as an int. Which is bad.

You can't access private members, and you shouldn't, that's why they are private.

int *p = (int *) &s;  
*p=10;

This means, you have an object s, of type sample, and it has a data member int a;. What you wrote works, because a happens to be at the beginning of the class.

There are some cases where the layout of a class is guaranteed, you can read about it here: