How to gain Access to member variables of a class using void pointer but Not Object

Question

I am trying to access member variables of a class without using object. please let me know how to go about.

class TestMem
{
    int a;
    int b;
public:
    Tes         


        

Answer 1

There totally is a way. C++ has member pointers, pointers relative to an object. They are defined by prefixing T:: to the * on the pointer type, and used by using the ->* or .* member pointer access operators. So yeah, it looks horrible :).

class T {
    int a, b;
public:
    typedef int T::* T_mem_ptr_to_int;

    static T_mem_ptr_to_int const a_ptr;
    static T_mem_ptr_to_int const b_ptr;
};

T::T_mem_ptr_to_int const T::a_ptr = &T::a;
T::T_mem_ptr_to_int const T::b_ptr = &T::b;

int weird_add(T* left, T* right) {
    return left->*T::a_ptr + right->*T::b_ptr;
}

This is used much more often for member function pointers, which look like Result (T::*ptr_name)(Arg1, Arg2, ...).

Answer 2

Short answer: You can't.

Long answer: You can, but it's highly implementation dependent. If you dump the memory you find at *p you'll see, somewhere around there, what you're looking for - a and b. But you will very likely also see some other stuff. What that stuff is, what it means, how big it is (and by implication where a and b actually live) is implementation dependent.

Answer 3

Simple answer: Don't do it.

There just can not be any situation where you can justify accessing like this. There just has to be a different solution.

Answer 4

I wanted to comment the answer provided by John Kugelman, being a new member didn't have enough reputation, hence posting it like an answer.

offsetof - is a C function used with structures where every member is a public, not sure whether we can refer the private variables as referred in the answer.

However the same can be achieved replacing the offsetof with a simple sizeof, ofcourse when we are sure of the type of the data members.

int vala = *reinterpret_cast<int *>(reinterpret_cast<char *>( ptr ) );
int valb = *reinterpret_cast<int *>(reinterpret_cast<char *>( ptr ) + sizeof ( int ) );

To my knowledge, you wouldn't be able to access. By the time you have assigned p, it doesn't refer to o1 here and p cannot replace pMem in (o1.*pMem)(), as p is not defined as function member to TestMem.

Answer 5

I came up with a solution but it's dirty:

class TestMem
{
public:
    int a;
    int b;
    TestMem(){}
    void TestMem1()
    {
        a = 10;
        b = 20;
    }
};

void* offset(void* ptr, ...)
{
  va_list ap;
  va_start(ap, ptr); // get 1st argument's address

  long i = va_arg(ap, long); // get next argument
  va_end(ap);

  return (char*)ptr + i;
}

void test()
{
  TestMem t;
  void* p = (TestMem*)&t;
  t.a = 8;
  t.b = 9;
  printf("%i\n", *(int*)offset(p, &TestMem::a));
  printf("%i\n", *(int*)offset(p, &TestMem::b));
}

Answer 6

The "right" way to do this is by using the offsetof() macro from <stddef.h>. Unfortunately offsetof() has some fairly draconian restrictions in C++:

Because of the extended functionality of structs in C++, in this language, the use of offsetof is restricted to "POD [plain old data] types", which for classes, more or less corresponds to the C concept of struct (although non-derived classes with only public non-virtual member functions and with no constructor and/or destructor would also qualify as POD).

So if you make a and b public and get rid of TestMem's constructor, you can write something like this to access a:

C++ style:

#include <cstddef>

int vala = *reinterpret_cast<int *>(reinterpret_cast<char *>(&o1)
           + offsetof(TestMem, a));

C style:

#include <stddef.h>

int vala = *(int *) ((char *) &o1 + offsetof(TestMem, a));

Notice that you need to use &o1 here, not p, which is a function pointer. The address of TestMem::TestMem1 won't have any relation to the locations of a and b. Class methods don't reside in memory anywhere near class member variables.

---

The "wrong" way is to just guess at where a and b are in memory. Most likely they are at offsets 0 and 4 from the start of o1, respectively. So this code would work most of the time:

int vala = *(int *) ((char *) &o1 + 0);
int valb = *(int *) ((char *) &o1 + 4);

There are a lot of assumptions here. This assumes that ints are 4 bytes and that there's no padding between a and b. On the other hand it doesn't have any of the restrictions from above: a and b don't need to be public, you can have a constructor, whatever.