C++. Error: void is not a pointer-to-object type

Bare bones example to reproduce the above error:

#include <iostream>
using namespace std;
int main() {
  int myint = 9;             //good
  void *pointer_to_void;     //good
  pointer_to_void = &myint;  //good

  cout << *pointer_to_void;  //error: 'void*' is not a pointer-to-object type
}

The above code is wrong because it is trying to dereference a pointer to a void. That's not allowed.

Now run the next code below, If you understand why the following code runs and the above code does not, you will be better equipped to understand what is going on under the hood.

#include <iostream>
using namespace std;
int main() {
    int myint = 9;
    void *pointer_to_void;
    int *pointer_to_int; 
    pointer_to_void = &myint;
    pointer_to_int = (int *) pointer_to_void;

    cout << *pointer_to_int;   //prints '9'
    return 0;
}

You are dereferencing the void * before casting it to a concrete type. You need to do it the other way around:

arguments vars = *(arguments *) (args);

This order is important, because the compiler doesn't know how to apply * to args (which is a void * and can't be dereferenced). Your (arguments *) tells it what to do, but it's too late, because the dereference has already occurred.

You have the * in the wrong place. So you're trying dereference the void*. Try this instead:

arguments vars = *(arguments *) (args);
std::cout << "\n" << vars.a << "\t" << vars.b << "\t" << vars.c << "\n";

Alternatively, you can do this: (which also avoids the copy-constructor - as mentioned in the comments)

arguments *vars = (arguments *) (args);
std::cout << "\n" << vars->a << "\t" << vars->b << "\t" << vars->c << "\n";