有沒有什麼辦法使這個模板專業化鏈接？

Question

我已經嘗試過以下的層次結構，它沒有鏈接。 c->Execute()的呼叫沒有被看到，因爲它似乎被Execute在Derived中掩蓋，並且它找不到適當的類型來使用。我會認爲如果出現問題，它將不得不在編譯時顯示。我已經使用GCC 4.8在線以及Windows上的borland編譯器進行了嘗試。這兩種情況下的錯誤消息都是相似的。東西的順序：

/tmp/ccrT5mNy.o:(.rodata._ZTV7DerivedI5TypeCE[_ZTV7DerivedI5TypeCE]+0x10): undefined reference to `Derived<TypeC>::Execute()' 
collect2: error: ld returned 1 exit status 

我會非常感謝任何指針。

#include <iostream> 

class BaseType 
{ 
public: 
    BaseType() {} 
    virtual void Execute() { std::cout << "BaseType execute..." << std::endl; }; 

protected: 
}; 

struct TypeA; 
struct TypeB; 
struct TypeC; 

class Derived2 : public BaseType 
{ 
public: 
    Derived2() : BaseType() {} 
    virtual void Execute() { std::cout << "Derived execute2" << std::endl; } 

protected: 
}; 


template <typename T> 
class Derived : public BaseType 
{ 
public: 
    Derived() : BaseType() {} 
    virtual void Execute(); 

protected: 
}; 

template <typename T> 
class GrandChild : public Derived<T> 
{ 
public: 
    GrandChild() : Derived<T>() {} 
    void Execute(); 
}; 

template<> 
void Derived<TypeA>::Execute() { std::cout << "Derived execute... TypeA" << std::endl; } 
template<> 
void Derived<TypeB>::Execute() { std::cout << "Derived execute... TypeB" << std::endl; } 
template<> 
void GrandChild<TypeC>::Execute() { std::cout << "Derived execute... TypeC" << std::endl; } 


int main() 
{ 
    BaseType* a = new Derived<TypeA>();  
    BaseType* b = new Derived<TypeB>();  
    BaseType* c = new GrandChild<TypeC>();  
    BaseType* d2 = new Derived2(); 
    a->Execute(); 
    b->Execute(); 
    c->Execute(); 
    d2->Execute(); 

    delete a; 
    delete b; 
    delete c; 
    delete d2; 
} 

Answer 1

Derived<TypeC>::Execute()必須定義，即使你從來沒有稱呼它，因爲你實例化類模板。嘗試添加一個空白的定義，或者一個，等等，程序應該按照你的期望鏈接和工作。