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

Question

I have a C++ program:

struct arguments
{
  int a, b, c;  
  arguments(): a(3), b(6), c(9) {}
};

class test_class{
  public:

    void *member_func(void *args){
      arguments vars = (arguments *) (*args); //error: void is not a 
                                              //pointer-to-object type

      std::cout << "\n" << vars.a << "\t" << vars.b << "\t" << vars.c << "\n";
    }
};

On compile it throws an error:

error: ‘void*’ is not a pointer-to-object type

Can someone explain what I am doing wrong to produce this error?

Solution 1

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.

Solution 2

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;
}

Solution 3

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";

Solution 4

The problem as bdonlan said is "dereferencing void* before casting".

I think this example would help:

#include <iostream>

using namespace std;

int main()
{



   void *sad;
   int s = 23;
   float d = 5.8;

   sad = &s;
   cout << *(int*) sad;//outputs 23//wrong: cout << *sad ;//wrong: cout << (int*) *sad;



   sad = &d;
   cout << *(float *) sad;//outputs 5.8//wrong: cout << *sad ;//wrong: cout << (float*) *sad;

   return 0;
}

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Comments

[visual_learner]

Leaving it as a pointer is probably better, since it avoids a copy constructor. However, the OP should make the pointer (both arguments * and void *) const.

[Billy ONeal]

But please use static_cast here -- you'd get a better error message that way...

[Matt Munson]

@Billy what does static_cast do?

[Billy ONeal]

@Matt: Same thing as a C style cast, except more limited. static_cast disallows removal of const, reinterpretation of a pointer type, and a few other nasty things. Generally speaking, anything that static_casts will work across platform/architectures. See section 5.2.9 of the C++ standard for more details (assuming C++03)