boost :: variant和polymorphism

Question

我希望从boost变量中获取指向基类的指针,如果我将orignally指针放到派生类上.有没有办法实现这一点.以下代码不起作用.

class A{ public: virtual ~A(){}}; class B : public A{};
typedef boost::variant<A*,B*> MyVar;
MyVar var = new B;
A* a = boost::get<A*> (var); // the following line throws exception

也许有人有想法怎么写我自己的get函数如果请求的类型是基类的存储类型的变种,这将考验,然后做相应的投

Answer 1

您可以使用operator()以下模板编写自己的访问者:

LIVE DEMO

#include <iostream>
#include <boost/variant.hpp>
#include <type_traits>

struct A { virtual ~A() {} virtual void foo() {} };
struct B : A { virtual void foo() { std::cout << "B::foo()" << std::endl; } };

template <typename T>
struct visitor : boost::static_visitor<T>
{
private:
    using Base = typename std::remove_pointer<
                        typename std::remove_cv<
                            typename std::remove_reference<T>::type
                        >::type
                    >::type;

    template <typename U>
    T get(U& u, std::true_type) const
    {
        return u;
    }

    template <typename U>
    T get(U& u, std::false_type) const
    {
        throw boost::bad_get{};
    }

public:
    template <typename U>
    T operator()(U& u) const
    {
        using Derived = typename std::remove_pointer<
                            typename std::remove_cv<
                                typename std::remove_reference<U>::type
                            >::type
                        >::type;

        using tag = std::integral_constant<bool
                         , (std::is_base_of<Base, Derived>::value
                           || std::is_same<Base, Derived>::value)
                           && std::is_convertible<U, T>::value>;

        return get(u, tag{});
    }
};

template <typename T, typename... Args>
T my_get(boost::variant<Args...>& var)
{
    return boost::apply_visitor(visitor<T>{}, var);
}

int main()
{    
    boost::variant<A*,B*> var = new B;

    A* a = my_get<A*>(var); // works!
    a->foo();

    B* b = my_get<B*>(var); // works!
    b->foo();
}

输出:

B::foo()
B::foo()

问答部分:

这个解决方案很奇怪!

不它不是.这正是Boost.Variant中访问者类的用途.类似的解决方案已经存在于最新版本的Boost.Variant中boost::polymorphic_get<T>.可悲的是,它是为其他目的而设计的,不能在这里使用.

Answer 2

您好，谢谢大家的回答和评论，我得出以下结论，它在编译时决定类型是否相互继承。它似乎有效，而且对我来说似乎更容易理解。

 #include <iostream>
 #include <boost/variant.hpp>
 #include <boost/type_traits.hpp>
 #include <boost/utility.hpp>

 using namespace boost::type_traits;


 struct A { virtual ~A() {} virtual void foo() {} };
 struct B : A { virtual void foo() { std::cout << "B::foo()" << std::endl; } };

  typedef boost::variant<B*,A*,C*> MyVar;


template <typename A,typename B> 
struct types_are_inheritance_related
{ 
 static const bool value=     
 ice_or<
 boost::is_base_of<A, B>::value,
 boost::is_base_of<B, A>::value
 >::value;    
};


 template<class Base>
 class get_visitor
: public boost::static_visitor<Base*> { public:


template<class T>
Base* operator()( T* t, typename boost::enable_if<types_are_inheritance_related<Base,T> >::type* dummy = 0)
{
   Base* b = dynamic_cast<Base*> ( t);
   return b;           
}  

template<class T>
Base* operator()( T* t, typename boost::disable_if<types_are_inheritance_related<Base,T> >::type* dummy = 0)
{       
   return 0;        
}   
};

template<class T>
T* get_var_value(MyVar& var)
{
   get_visitor<T> visitor;
   T* aa= var.apply_visitor(visitor);
   return aa;
}

int main()
{    
 MyVar var = new B;

 A* a = get_var_value<A*>(var); // works!
 a->foo();

 B* b = get_var_value<B*>(var); // works!
 b->foo();
}