从功能结构成员获取功能地址

Question

I'm trying to get function addresses which are hidden behind structures. Unfortunately, the basic C++ conversion doesn't work, so I used instead.

void * lpfunction;
lpfunction = scanf; //OK
lpfunction = MessageBoxA; //OK

I made a simple structure :

struct FOO{

    void PRINT(void){printf("bla bla bla");}

    void SETA(int){} //nothing you can see
    void SETB(int){} //nothing you can see

    int GETA(void){} //nothing you can see
    int GETB(void){} //nothing you can see
};
///////////////////////////////////////////
void *lpFunction = FOO::PRINT;

And the compiling error :

error C2440: 'initializing' : 
cannot convert from 'void (__thiscall FOO::*)(void)' to 'void *'

2. Is getting function member addresses impossible?

Then, I made a template function which is able to convert a function member to address. Then I will call it by assembly. It should be something like this:

template <class F,void (F::*Function)()>  
void * GetFunctionAddress() {

    union ADDRESS  
    { 
        void (F::*func)();  
        void * lpdata;  
    }address_data;  

    address_data.func = Function;  
    return address_data.lpdata; //Address found!!!  

}  

And here is the code :

int main()
{
    void * address = GetFunctionAddress<FOO,&FOO::PRINT>();

    FOO number;
    number.PRINT(); //Template call
    void * lpdata = &number;

printf("Done.\n");
system("pause");
return 0;
}

But, I see it is extremely specific . It looks like LOCK - KEY , and I have to make a new template for every set of argument types.

Original (OK) :

void PRINT(); //void FOO::PRINT();

Modify a bit :

void PRINT(int); //void FOO::PRINT(int);

Immediately with old template code the compiler shows :

//void (F::*func)();
//address_data.func = Function;

They are only addresses.

69:       address_data.func = Function;
00420328  mov  dword ptr [ebp-4],offset @ILT+2940(FOO::PRINT) (00401b81)

...

EDIT3 : I know the better solution :

void(NUMBER::*address_PRINT)(void) = FOO::PRINT;
int(NUMBER::*address_GETA)(void) = FOO::GETA;
int(NUMBER::*address_GETB)(void) = FOO::GETB;
void(NUMBER::*address_SETA)(int) = FOO::SETA;
void(NUMBER::*address_SETA)(int) = FOO::SETB;

It's much better than template . And by the way I want to achieve the goal :

<special_definition> lpfunction;
lpfunction = FOO::PRINT; //OK
lpfunction = FOO::GETA; //OK
lpfunction = FOO::GETB; //OK
lpfunction = FOO::SETA; //OK
lpfunction = FOO::SETB; //OK

Is this possible?

Answer 1

Pointers to member functions are nothing like pointers to global functions or static member functions. There are many reasons for this, but I'm not sure how much you know about how C++ works, and so I'm not sure what reasons will make sense.

I do know that what you are trying in assembly simply won't work in the general case. It seems like you have a fundamental misunderstanding about the purpose of member functions and function pointers.

The thing is, you are doing some things that you would generally not do in C++. You don't generally build up tables of function pointers in C++ because the things you would use that sort of thing for are what virtual functions are for.

If you are determined to use this approach, I would suggest you not use C++ at all, and only use C.

To prove these pointer types are completely incompatible, here is a program for you:

#include <cstdio>

struct Foo {
   int a;
   int b;
   int addThem() { return a + b; }
};

struct Bar {
   int c;
   int d;
   int addThemAll() { return c + d; }
};

struct Qux : public Foo, public Bar {
   int e;
   int addAllTheThings() { return Foo::addThem() + Bar::addThemAll() + e; }
};

int addThemGlobal(Foo *foo)
{
   return foo->a + foo->b;
}

int main()
{
   int (Qux::*func)();

   func = &Bar::addThemAll;
   printf("sizeof(Foo::addThem) == %u\n", sizeof(&Foo::addThem));
   printf("sizeof(Bar::addThemAll) == %u\n", sizeof(&Bar::addThemAll));
   printf("sizeof(Qux::addAllTheThings) == %u\n", sizeof(&Qux::addAllTheThings));
   printf("sizeof(func) == %u\n", sizeof(func));
   printf("sizeof(addThemGlobal) == %u\n", sizeof(&addThemGlobal));
   printf("sizeof(void *) == %u\n", sizeof(void *));
   return 0;
}

On my system this program yields these results:

$ /tmp/a.out 
sizeof(Foo::addThem) == 16
sizeof(Bar::addThemAll) == 16
sizeof(Qux::addAllTheThings) == 16
sizeof(func) == 16
sizeof(addThemGlobal) == 8
sizeof(void *) == 8

Notice how the member function pointer is 16 bytes long. It won't fit into a void * . It isn't a pointer in the normal sense. Your code and union work purely by accident.

The reason for this is that a member function pointer often needs extra data stored in it related to fixing up the object pointer it's passed in order to be correct for the function that's called. In my example, when called Bar::addThemAll on a Qux object (which is perfectly valid because of inheritance) the pointer to the Qux object needs to be adjusted to point at the Bar sub-object before the function is called. So Qux::* s to member functions must have this adjustment encoded in them. After all, saying func = &Qux::addAllTheThings is perfectly valid, and if that function were called no pointer adjustment would be necessary. So the pointer adjustment is a part of the function pointer's value.

And that's just an example. Compilers are permitted to implement member function pointers in any way they see fit (within certain constraints). Many compilers (like the GNU C++ compiler on a 64-bit platform like I was using) will implement them in a way that do not permit any member function pointer to be treated as at all equivalent to normal function pointers.

There are ways to deal with this. The swiss-army knife of dealing with member function pointers is the ::std::function template in C++11 or C++ TR1.

An example:

 #include <functional>

 // .... inside main
    ::std::function<int(Qux *)> funcob = func;

funcob can point at absolutely anything that can be called like a function and needs a Qux * . Member functions, global functions, static member functions, functors... funcob can point at it.

That example only works on a C++11 compiler though. But if your compiler is reasonably recent, but still not a C++11 compiler, this may work instead:

 #include <tr1/functional>

 // .... inside main
    ::std::tr1::function<int(Qux *)> funcob = func;

If worse comes to worse, you can use the Boost libraries, which is where this whole concept came from.

But I would rethink your design. I suspect that you will get a lot more milage out of having a well thought out inheritance hierarchy and using virtual functions than you will out of whatever it is you're doing now. With an interpreter I would have a top level abstract 'expression' class that is an abstract class for anything that can be evaluated. I would give it a virtual evaluate method. Then you can derive classes for different syntax elements like an addition expression a variable or a constant. Each of them will overload the evaluate method for their specific case. Then you can build up expression trees.

Not knowing details though, that's just a vague suggestion about your design.

Answer 2

Here is a clean solution. By means of a template wrap your member function into a static member function. Then you can convert it to whatever pointer you want:

template<class F, void (F::*funct)()>
struct Helper: public T {
  static void static_f(F *obj) {
    ((*obj).*funct)();
  };
};

struct T {
  void f() {
  }
};

int main() {
  void (*ptr)(T*);
  ptr = &(Helper<T,&T::f>::static_f);
}

Answer 3

It seems that you need to convert a pointer to a member function to a void *. I presume you want to give that pointer as a "user data" to some library function and then you will get back your pointer and want to use it on some given object.

If this is the case a reinterpret_cast<void *>(...) could be the right thing... I assume that the library receiving the pointer is not using it.