有没有转发方法的通用方法? [英] Is there a generic way to forward constructor arguments?

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问题描述

我下面有一个有效的 Cloneable / CloneableImpl 类对。只要我有从子级到父级的默认构造函数,它就可以工作。

I have a working Cloneable/CloneableImpl class pair below. It does its job as long as I have default constructors from child to parent.

假设将Animal的构造函数修改为 Animal(std :: string const& ; name),它需要从子类构造函数中传递名称。

Suppose Animal's constructor is modified to Animal( std::string const& name ), which needs the name to be passed up from the children class constructors.

如何在保持 Cloneable / CloneableImpl 通用性的同时将这一要求纳入结构中?

How can I incorporate this requirement into the structure while keeping Cloneable/CloneableImpl generic?

换句话说,我需要能够将所有构造函数参数从Lion,Tiger转发到Animal。有没有一种方法可以在C ++ 11中以通用方式执行此操作?

如果不可能,如何将这些模板进行重组以保持通用性

If it's not possible, how can these templates be restructured to remain generic while allowing for the constructor requirement?

代码

template<typename P>
struct Cloneable
{
    virtual P* clone() const = 0;
};

template<typename T,typename P>
struct CloneableImpl :
    public P
{
    virtual P* clone() const 
    {
        return new T( dynamic_cast<T const&>(*this));
    }
};

// ----------------------------------------------------------------------------

struct Animal :
    public Cloneable<Animal>
{ 
};

struct Lion  : 
    public CloneableImpl<Lion,Animal> 
{ 
};

struct Tiger : 
    public CloneableImpl<Tiger,Animal> 
{ 
};

int
main( int argv, char* argc[] )
{
    Animal* x = new Lion;
    Animal* y = x->clone();

    // we want to do this without hard-coding in template classes
    // Animal* z = new Lion( "Samba" );  
}


推荐答案

跟随@ Cheersandhth.-Alf和@R。 Martinho Fernandes在对OP的评论中的建议,以调查完美转发。我研究了一下,并提出了这个建议,似乎可行。

Following @Cheersandhth.-Alf and @R. Martinho Fernandes' suggestions in the Comments to the OP to look into perfect forwarding. I researched a bit and came up with this, which seems to work.

谢谢大家!

CODE

#include <string>
#include <iostream>

template<typename P>
struct Cloneable
{
    virtual P* clone() const = 0;
};

template<typename T,typename P>
struct CloneableImpl :
    public P
{
    template<typename... Args>
    CloneableImpl( Args&&... args ) 
       : P(std::forward<Args>(args)...)
  {  }

    virtual P* clone() const 
    {
        return new T( dynamic_cast<T const&>(*this));
    }
};

// ----------------------------------------------------------------------------

struct Animal :
    public Cloneable<Animal>
{ 
    Animal( std::string const& name ) : m_name( name ) { }

    std::string  m_name;
};

struct Lion  : 
    public CloneableImpl<Lion,Animal> 
{ 
    template<typename... Args>
    Lion( Args&&... args ) 
       : CloneableImpl<Lion,Animal>(std::forward<Args>(args)...)
  {  }
};

struct Tiger : 
    public CloneableImpl<Tiger,Animal> 
{ 
};

int
main( int argv, char* argc[] )
{
    Animal* x = new Lion( "Samba" );
    Animal* y = x->clone();

    std::cerr << y->m_name << std::endl;
} 

这篇关于有没有转发方法的通用方法?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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