如何處理相互依賴且具有模板成員的類？

Question

這部分與this SO question有關。

我有兩個班，他們兩個都是模板，例如：

class Base 
{ 
public: 
    template< class T > void operator=(T other) 
    { 
     //irrelevant 
    } 

    Derived toDerived() 
    { 
     Derived d; 
     //do stuff; 
     return d; 
    } 
}; 

class Derived: public Base 
{ 
public: 
    template< class T > void foo(T other) 
    { 
     //do stuff 
    } 
}; 

正如你所看到的，兩者都模板和Base類函數內我需要創建的Derived一個實例。當然，現在我得到一個錯誤Derived does not name a type。不幸的是，我不能僅僅提前宣佈Derived，因爲它會導致另一個錯誤variable 'Derived d ' has initializer but incomplete type。

從上面提到的SO問題我明白，編譯器需要知道所有模板參數能夠正確地轉發聲明它。但顯而易見，我不能僅僅移動Derived聲明，因爲它會導致完全相同的問題，反之亦然。

有沒有辦法做到這一點？

Answer 1

// Declare, but do not define 
class Derived; 

class Base { 
public:  
    // Declare, but do not define 
    // at this point Derived must be declared to be able 
    // to use it in the return type 
    Derived toDerived(); 
}; 

// Define 
class Derived: public Base { 
    // Rest of definition 
}; 

// At this point Derived is defined 

// Define 
Derived Base::toDerived() 
{ 
    // Implementation goes here 
} 

Answer 2

此問題與模板無關。你可以只使用前向申明的Derived編譯的Base::toDerived()申報和移動功能定義 取決於後DerivedDerived定義：

// Forward declaration of Derived 
class Derived; 

// Definition of Base 
class Base 
{ 
public: 
    // Base::toDerived() declaration only 
    Derived toDerived(); 
}; 

// Definition of Derived 
class Derived: public Base 
{ 
public: 
... 
}; 

// Base::toDerived() definition 
inline Derived Base::toDerived() 
{ 
    Derived d; 
    // Do dirty things 
    return d; 
} 

Answer 3

你可以做

class Derived; 

class Base 
{ 
public: 
    template< class T > void operator=(T other) 
    { 
     //irrelevant 
    } 

    Derived toDerived(); 
}; 

class Derived: public Base 
{ 
public: 
    template< class T > void foo(T other) 
    { 
     //do stuff 
    } 
}; 

Derived Base::toDerived() 
{ 
    Derived d; 
    //do stuff; 
    return d; 
} 

正如你可以看到它與模板無關。

另外，這種設計根本感覺不對。