Test whether a class is polymorphic

Question

Hello Group,

We have a sub-project 'commonUtils' that has many generic code-snippets used across the parent project. One such interesting stuff i saw was :-

/*********************************************************************
If T is polymorphic, the compiler is required to evaluate the typeid 
stuff at runtime, and answer will be true.  If T is non-polymorphic, 
the compiler is required to evaluate the typeid stuff at compile time, 
whence answer will remain false
*********************************************************************/
template <class T> 
bool isPolymorphic() { 
   bool answer=false; 
   typeid(answer=true,T()); 
   return answer; 
}

I believed the comment and thought that it is quite an interesting template though it is not used across the project. I tried using it like this just for curiosity ...

class PolyBase {
public:
   virtual ~PBase(){}
};

class NPloyBase {
public:
   ~NBase(){}
};


if (isPolymorphic<PolyBase>())
  std::cout<<"PBase = Polymorphic\n";
if (isPolymorphic<NPolyBase>())
  std::cout<<"NBase = Also Polymorphic\n";

But none of those ever returns true. MSVC 2005 gives no warnings but Comeau warns typeid expression has no effect. Section 5.2.8 in the C++ standard does not say anything like what the comment says i.e. typeid is is evaluated at compile time for non-polymorphic types and at runtime for polymorphic types.

1) So i guess the comment is misleading/plain-wrong or since the author of this code is quite a senior C++ programmer, am i missing something?

2) OTOH, I am wondering if we can test whether a class is polymorphic(has at least one virtual function) using some technique?

3) When would one want to know if a class is polymorphic? Wild guess; to get the start-address of a class by using dynamic_cast<void*>(T) (as dynamic_cast works only on polymorphic classes).

Awaiting your opinions.

Thanks in advance,

Answer 1

class PolyBase {
public:   
    virtual ~PolyBase(){}
};

class NPolyBase {
public:
    ~NPolyBase(){}
};

template<class T>
struct IsPolymorphic
{
    struct Derived : T {
     virtual ~Derived();
    };
    enum  { value = sizeof(Derived)==sizeof(T) };
};


void ff()
{
    std::cout << IsPolymorphic<PolyBase >::value << std::endl;
    std::cout << IsPolymorphic<NPolyBase>::value << std::endl;
}

Answer 2

I cannot imagine any possible way how that typeid could be used to check that type is polymorphic. It cannot even be used to assert that it is, since typeid will work on any type. Boost has an implementation here. As for why it might be necessary -- one case I know is the Boost.Serialization library. If you are saving non-polymorphic type, then you can just save it. If saving polymorphic one, you have to gets its dynamic type using typeid, and then invoke serialization method for that type (looking it up in some table).

Update: it appears I am actually wrong. Consider this variant:

template <class T> 
bool isPolymorphic() { 
    bool answer=false;
    T *t = new T();
    typeid(answer=true,*t); 
    delete t;
    return answer; 
}

This actually does work as name suggests, exactly per comment in your original code snippet. The expression inside typeid is not evaluated if it "does not designate an lvalue of polymorphic class type" (std 3.2/2). So, in the case above, if T is not polymorphic, the typeid expression is not evaluated. If T is polymorphic, then *t is indeed lvalue of polymorphic type, so entire expression has to be evaluated.

Now, your original example is still wrong :-). It used T(), not *t. And T() create rvalue (std 3.10/6). So, it still yields an expression that is not "lvalue of polymorphic class".

That's fairly interesting trick. On the other hand, its practical value is somewhat limited -- because while boost::is_polymorphic gives you a compile-time constant, this one gives you a run-time value, so you cannot instantiate different code for polymorphic and non-polymorphic types.