解决模板类之间的循环依赖关系

您需要做的是将类声明与实现分离开来。比如说

template <class T> class Foo : public Base
{
    public:
    Base* convert(ID) const;
}

template <class T> class Bar : public Base
{
    public:
    Base* convert(ID) const;
}

template <class T> Base* Foo<T>::convert(ID) const {return new Bar<T>;}
template <class T> Base* Bar<T>::convert(ID) const {return new Foo<T>;}

这样,在定义函数时就有了完整的类定义。

(已更新)

#if !defined(BAR_H_INCLUDED)
#define BAR_H_INCLUDED

template <class T>
class Foo;

template <class T>
class Bar
{
    /// Declarations, no implementations.
}    

#include "Foo.h"

template <class T>
Base* Bar<T>::Convert() {  /* implementation here... */ }
#endif

template <class T>
class X;

是非常好的模板类转发声明。

t1.hh公司

#ifndef S_SIGNATURE
#define S_SIGNATURE

struct G; // forward declaration

template<typename T>
struct S {
  void s_method(G &);
};

#include "t2.hh" // now we only need G's member declarations

template<typename T>
void S<T>::s_method(G&g) { g.g_method(*this); }

#endif

t2.hh小时

#ifndef G_SIGNATURE
#define G_SIGNATURE

template<typename T>
struct S; // forward declaration

struct G {
  template<typename T>
  void g_method(S<T>&);
};

#include "t1.hh" // now we only need S' member declarations

template<typename T>
void G::g_method(S<T>& s) { s.s_method(*this); }

#endif

t、 抄送

#include "t1.hh"
#include "t2.hh"

S<int> s;
G g;

int main(int argc,char**argv) {
  g.g_method(s); // instantiation of G::g_method<int>(S<int>&)
}