Is there a way to forward declare covariance?

Question

Suppose I have these abstract classes Foo and Bar:

class Foo;
class Bar;

class Foo
{
public:
  virtual Bar* bar() = 0;
};

class Bar
{         


        

Answer 1

Doesn't static polymorphism solve your problem? Feeding the base class with the derived class through template argument? So the base class will know the derivative Type and declare a proper virtual?

Answer 2

How about this.

template <class BarType>
class Foo
{
public:
    virtual BarType* bar() = 0;
};

template <class FooType>
class Bar
{
public:
    virtual FooType* foo() = 0;
};

class ConcreteBar;
class ConcreteFoo : public Foo<ConcreteBar>
{
public:
    ConcreteBar* bar();
};

class ConcreteBar : public Bar<ConcreteFoo>
{
public:
    ConcreteFoo* foo();
};

Answer 3

You can fake it quite easily, but you lose the static type checking. If you replace the dynamic_casts by static_casts, you have what the compiler is using internally, but you have no dynamic nor static type check:

class Foo;
class Bar;

class Foo
{
public:
  Bar* bar();
protected:
  virtual Bar* doBar();
};

class Bar;
{
public:
  Foo* foo();
public:
  virtual Foo* doFoo();
};

inline Bar* Foo::bar() { return doBar(); }
inline Foo* Bar::foo() { return doFoo(); }

class ConcreteFoo;
class ConcreteBar;
class ConcreteFoo : public Foo
{
public:
  ConcreteBar* bar();
protected:
  Bar* doBar();
};

class ConcreteBar : public Bar
{
public:
   ConcreteFoo* foo();
public:
   Foo* doFoo();
};

inline ConcreteBar* ConcreteFoo::bar() { return &dynamic_cast<ConcreteBar&>(*doBar()); }
inline ConcreteFoo* ConcreteBar::foo() { return &dynamic_cast<ConcreteFoo&>(*doFoo()); }

Answer 4

Covariance is based on inheritance diagram, so since you cannot declare

class ConcreteBar : public Bar;

hence no way to tell compiler about covariance.

But you can do it with help of templates, declare ConcretFoo::bar as template and later bounding allows you solve this problem