Re: r H2 deduce deduce template argument of a template class inheriting from a non template base?

From:
SG <s.gesemann@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 6 Mar 2009 13:44:21 -0800 (PST)
Message-ID:
<7d8a04f7-eb10-4937-b5b3-033a2f7dbf73@d19g2000yqb.googlegroups.com>
On 6 Mrz., 21:11, nguillot <nicolas.guil...@gmail.com> wrote:

Let's say I've got a template class inheriting from a non template
class:

struct SData
{
    virtual ~SData() {}
};


Are you sure you don't need some more functions? Perhaps something
like this:

  struct SData
  {
      virtual ~SData() {}
      virtual void print_on(std::ostream&) const = 0;
  };

That's what virtual functions are for. ;-)

template <class T>
struct SDataContainer : public SData
{
    SDataContainer(const T& o) : m_data(o) {}
    T m_data;
};


You could override "print_on" here.

 void print(SData* pData)
 {
     SDataContainer<int>* pi = dynamic_cast<SDataContainer<int>*>
(pData);
     if (pi != 0)
     {
         cout << "printing integer " << pi->m_data << endl;
         return;
     }

     SDataContainer<bool>* pb = dynamic_cast<SDataContainer<bool>*>
(pData);
     if (pb != 0)
     {
         cout << "printing bool " << (pb->m_data ? "true" :
"false") << endl;
         return;
     }


BTW: Is there a reason why you put this function inside a class or is
this only a habit you imported from Java / C# for no reason?

You're right. That looks ugly. You could avoid this with a virtual
function like SData::print_on :

  void print(SData* pData)
  {
    pData->print_on(cout);
  }

Now, the question: is there a way to avoid the dynamic_cast and deduce
the template argument thanks to template method specialization.
I would like to write something like that: [...]

    template <class T>
    void print(SDataContainer<T>* pData)
    {
        cout << "data type not handled" << endl;
    }

    template <>
    void print(SDataContainer<int>* pData)
    {
        cout << "printing integer " << pData->m_data << endl;
    }


(1) Specializing function templates is discouraged.
(2) Your syntax seems wrong: Shouldn't it be "print<int>" ?
(3) There's a big difference between the static type and the
    runtime (most derived) type. The first one is known at
    compile-time the 2nd one can only be determined at
    run-time (in general). So, the compiler isn't able to select
    the function with the right runtime type at compile-time.

That's what virtual functions are for.

Cheers!
SG

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