从基指针向模板化派生类型的向下转换

时间:2012-03-05 02:16:08

标签: c++ inheritance metaprogramming dynamic-cast downcast

我有以下层次结构:

class base
{
public:
   virtual ~base(){}
   virtual void foo() {}
};

template <typename T>
class derived1 : public base
{
   virtual void foo() {};
};

template <typename T>
class derived2 : public base
{
   virtual void foo() {};
};

现在给出一个指向base的指针,我想知道是否 底层是derived1或derived2。问题是 derived1和derived2都可以专门用于许多不同的 类型,使用dynamic_cast来测试向下转换需要 模板类型要知道。我最终得到了凌乱,不稳定和不完整的代码:

base* b = new derived1<int>();

if (dynamic_cast<derived1<int>*> ||
    dynamic_cast<derived1<unsigned int>*> ||
    dynamic_cast<derived1<double>*>)
  std::cout << "is derived1";
else if (dynamic_cast<derived2<int>*> ||
    dynamic_cast<derived2<unsigned int>*> ||
    dynamic_cast<derived2<double>*>)
  std::cout << "is derived2";

有没有更好的方法,可以处理任何类型的专业化?

4 个答案:

答案 0 :(得分:9)

将依赖于类型的逻辑移动到类型中。

而不是:

if (dynamic_cast<derived1<int>*>(b) ||
    dynamic_cast<derived1<unsigned int>*>(b) ||
    dynamic_cast<derived1<double>*>(b))
  std::cout << "is derived1";
else if (dynamic_cast<derived2<int>*>(b) ||
    dynamic_cast<derived2<unsigned int>*>(b) ||
    dynamic_cast<derived2<double>*>(b))
  std::cout << "is derived2";

virtual print_name() const函数添加到base,然后执行:

void example() {
    std::unique_ptr<base> b(new derived1<int>());
    b->print_name();
}
class base
{
public:
   ~base(){}
   virtual void foo() {}
   virtual void print_name() const = 0;
};

template <typename T>
class derived1 : public base
{
   virtual void foo() {}
   virtual void print_name() const {
       std::cout << "is derived1";
   }
};

template <typename T>
class derived2 : public base
{
   virtual void foo() {}
   virtual void print_name() const {
       std::cout << "is derived2";
   }
};

答案 1 :(得分:7)

basederived1derived2之间插入一个非模板类:

class base
{
public:
   virtual ~base() {}  // **NOTE** Should be virtual
   virtual void foo() {}
};

class derived1_base : public base
{
};

template <typename T>
class derived1 : public derived1_base
{
public:
   virtual void foo() {}
};

class derived2_base : public base
{
};

template <typename T>
class derived2 : public derived2_base
{
public:
   virtual void foo() {}
};

在评论中,您提到:

  

[我想]为每个人调用一个特定的功能 - 顺便说一下,还有更多功能   derived1和derived2

将该(虚拟)功能添加到derived1_base,您甚至不需要再知道T

if (dynamic_cast<derived1_base*>(foo))
{
  std::cout << "is derived1";
  dynamic_cast<derived1_base*>(foo)->specific_derived1_function();
}
else if (dynamic_cast<derived2_base*>(foo))
{
  std::cout << "is derived2";
  dynamic_cast<derived2_base*>(foo)->specific_derived2_function();
}

注意:我会考虑dynamic_cast<> 代码异味的列表,我建议您重新考虑一下您的方法。

答案 2 :(得分:3)

您可以添加虚拟方法来执行某种类型的元类型检查:

class base
{
public:
    ~base(){}
    virtual void foo() {}
    virtual bool isa(const char* type_to_test){
          return strcmp(type_to_test,"base")==0;}
};

template <typename T>
class derived1 : public base
{
   virtual void foo() {};
   virtual bool isa(const char* type_to_test){
   return strcmp(type_to_test,"derived1")==0;}
};

答案 3 :(得分:2)

解决方案1:再添加一个虚函数:

enum DerivedType
{
    One,
    Two,
    ...
};

class base 
{ 
public: 
   ~base(){} 
   virtual void foo() {}
   virtual DerivedType GetType() = 0;
}; 

template <typename T> 
class derived1 : public base 
{ 
   virtual void foo() {};
   virtual DerivedType GetType() { return One; }
}; 

template <typename T> 
class derived2 : public base 
{ 
   virtual void foo() {};
    virtual DerivedType GetType() { return Two; }
}; 

解决方案2:使用标记类:

class Base
{
public:
    virtual ~Base() { }
};

class Derived1Tag
{ };

class Derived2Tag
{ };

template <class T>
class Derived1 : public Base, public Derived1Tag
{ };

template <class T>
class Derived2 : public Base, public Derived2Tag
{ };


int main(int argc, char** argv)
{
    Derived1<int> d1;
    Derived2<int> d2;

    cout << dynamic_cast<Derived1Tag*>((Base*)&d1) << endl;
    cout << dynamic_cast<Derived1Tag*>((Base*)&d2) << endl;

    return 0;
}