Question

我尝试了以下层次结构但它没有链接。没有看到对c->Execute()的调用，因为它似乎被Execute中的Derived屏蔽，并且它找不到足够的类型来使用。我原本以为如果出现问题，就必须在汇编时显示。我已尝试使用GCC 4.8在线和在Windows上使用borland编译器。两种情况下的错误消息都类似。按顺序排列：

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

我会非常感谢任何指示。

#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

即使您从未调用过，也必须定义

Derived<TypeC>::Execute()，因为您正在实例化类模板。尝试添加一个空的定义，或throw s等的定义，程序应链接并按预期工作。

是否有任何方法使这个模板专业化链接？

1 个答案: