开发者

How to determine if a C++ class has a vtable?

A friend of mine sent me the following challenge earlier today:

Given the following code, propose an implementation of OBJECT_HAS_VTABLE so the program prints AnObject has a vtable = 0, AnObjectWithVTable has a vtable = 1.

class AnObject
{
    int m_a;
    void DoSomething() {}

public: 
    AnObject() {m_a = 0;}
};

class AnObjectWithVTable
{
    int m_b;
    virtual void DoStuff() { }

public: 
    AnObjectWithVTable() {m_b = 0;}
};

void main()
{
    printf("AnObject has a vtable = %i, AnObjectWithVTable has a vtable = %i\n",
           OBJECT_HAS_VTABLE(AnObject),
           OBJECT_HAS_VTABLE(AnObjectWithVTable));
}

I've came up with the following solution which I think is decent enough:

template <typename T>
bool objectHasVtable()
{
    class __derived : public T {};
    T t;
    __开发者_开发技巧derived d;

    void *vptrT=*((void **)&t);
    void *vptrDerived=*((void **)&d);

    return vptrT != vptrDerived;
}

#define OBJECT_HAS_VTABLE(T) objectHasVtable<T>()

Is there a better solution to this problem?

Edit

The solution doesn't have to be generic across all compilers. It can work on gcc, g++, MSVC... Just specify for which compiler your solution is known to be valid. Mine is for MSVC 2010.


The standard method is to use std::is_polymorphic from C++11/C++03 TR1/Boost to determine if a class (and its bases) contain any virtual members.

#include <type_traits>
#define OBJECT_HAS_VTABLE(T) (std::is_polymorphic<T>::value)


For completeness sake, here's the answer my buddy just sent me. From the look of it, it's probably similar to how TR1 does it (though I haven't looked at the code myself).

template<class T>
class HasVTable
{
public :
    class Derived : public T
    {
        virtual void _force_the_vtable(){}
    };
    enum { Value = (sizeof(T) == sizeof(Derived)) };
};

#define OBJECT_HAS_VTABLE(type) HasVTable<type>::Value


You can use the following properties of C++:

  1. dynamic_cast fails at compile time if the argument is not a polymorphic class. Such a failure can be used with SFINAE.
  2. dynamic_cast<void*> is a valid cast that returns the address of the complete polymorpic object.

Hence, in C++11:

#include <iostream>
#include <type_traits>

template<class T>
auto is_polymorphic2_test(T* p) -> decltype(dynamic_cast<void*>(p), std::true_type{});

template<class T>
auto is_polymorphic2_test(...) -> std::false_type;

template<class T>
using is_polymorphic2 = decltype(is_polymorphic2_test<T>(static_cast<T*>(0)));

struct A {};
struct B { virtual ~B(); };

int main() {
    std::cout << is_polymorphic2<A>::value << '\n'; // Outputs 0.
    std::cout << is_polymorphic2<B>::value << '\n'; // Outputs 1.
}
0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜