Inheritance issues with template classes

Question

I'm having issues with a very strange error in some code I wrote. The basic idea behind the code can be trivialised in the following example:


template <class f, class g> class Ptr;

template <class a, class b, class c = Ptr<a,b> >
class Base
{
public:
    Base(){};
};

template <class d, class e>
class Derived : public Base <d,e>
{
public:
    Derived(){};
};

template <class f, class g>
class Ptr
{
public:
    Ptr(){};
    Ptr(Base<f,g,Ptr<f,g> >* a){};
};

typedef Derived<double,double> DDerived;

int main()
{
    Base<int,int> b = Base<int,int>();
    Derived<double,double> d = Derived<double,double>();
    DDerived dd = DDerived();
    Ptr<double,double> p(&dd);
    return 1;
}

The basic idea is that pointers are replaced by the Ptr class (This will eventually be used in an MPI setting so standard pointers are effectively useless). The pointers are designed to 'point' at the base class, and so can point at any inherited class (as demonstrated in the example).

Can anyone think of any reason this might not work in a non-trivial case (But a case where the object architecture remains identical).

The error that is occurring in the main case is as follows:


void function()
{
  vector nVector(1);  // cut down for simplicity
  nVector[0].SetId(1); // To ensure the node is instantiated correctly
  Ptr temp(&nVector[1]);
};

This code produces the (slightly extended version of the) error when compiled with MPICXX:
no matching function for call to Ptr<double, double>::Ptr(Derived<double, double>*)
candidates are . . . (Some removed for simplicity's sake)
Ptr<f, g>::Ptr(Base<f, g, Ptr<f, g> >*) [with f = double, g = double]

Cheers, Ed

EDITED (Detailing the error a little better, added info on the compiler)

+1 for making the effort of reducing the code to its simplest form. — cletus
– cletus, Commented Feb 25, 2009 at 13:39
Your error-causing code seems to have been eaten by SO... but I'm not having issues with std::vector<Base<double, double> > (VC++9), so it may be some weirdness with the underlying C++ compiler (g++?). — Simon Buchan
– Simon Buchan, Commented Feb 27, 2009 at 5:27
Not a compiler error as far as I can tell -> the code I gave compiles fine but the higher-level version doesn't. — Ed James
– Ed James, Commented Feb 27, 2009 at 16:12

Ed James · Accepted Answer · 2012-10-08 21:07:51Z

1

Unfortunately, I was being slightly stupid and had forgotten to put my Ptr class in the same namespace as the Base and Derived classes.

That, I guess, would be why it wasn't working ! =]

edited Oct 8, 2012 at 21:07

answered Feb 27, 2009 at 17:09

Ed James

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Comments

Evgeny Lazin · Accepted Answer · 2009-02-25 13:50:22Z

0

Try to use template template parameter instead of default value.

template <class a, class b, template <class aa, class bb> class c = Ptr >
class Base
{
typedef Ptr<a, b> pointer;
public:
    Base(){};
};

answered Feb 25, 2009 at 13:50

Evgeny Lazin

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1 Comment

Ed James Over a year ago

Erm. . . This doesn't compile and I'm not really sure what you're getting at? You seem to be trying to define a class inside a template argument.

David Allan Finch · Accepted Answer · 2009-02-25 16:35:42Z

0

Ok what about this:

class BasePtr
{
public:
    virtual void* obj() = 0;
};

template <class T>
class Ptr :: public BasePtr
{
public:
    Ptr() : ptr(0) {};
    Ptr(T* a) : ptr(a) {};
    virtual T* obj() { return ptr; }
protected:
    T* ptr;
};

Use BasePtr in the Base class but pass it the correct template object when Base needs to use it.

template <class a, class b >
class Base
{
public:
    Base(){};
    void set_ptr( BasePtr* );
};

DDerived dd = DDerived();
Ptr<double,double> p(&dd);
dd.set_ptr( p );

I still do not quite understand your problem but I hope that helps.

edited Feb 25, 2009 at 16:35

answered Feb 25, 2009 at 13:48

David Allan Finch

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3 Comments

Ed James Over a year ago

Unfortunately impossible in the more complex scenario, as it ends up with a recursive template definition: template <class A, class B, class c = Ptr<A,B,Ptr<A,B . . . > > This is not an option, although it was the original design.

Ed James Over a year ago

Second revision: I understand your idea but it's not an option, the template argument is for a typedef, and so cannot be allocated after instantiation with a method. The templated argument is used to override the standard Ptr object with an MpiPtr when the code is used in a distributed fashion.

David Allan Finch Over a year ago

Why can't MpiPtr derived from BasePtr?

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