C ++通过调用基类优雅地克隆派生类 [英] C++ elegantly clone derived class by calling base class

问题描述

我需要克隆一个派生类，只给出一个引用或指向基类的指针。下面的代码做的工作，但似乎不优雅，因为我把样板代码到许多派生类C，D，E是兄弟的B（未显示），只是调用每个的默认副本构造函数。是不是默认的复制构造函数是为了，如果只有它可以是虚拟？

I have a need to clone a derived class given only a reference or pointer to the base class. The following code does the job, but doesn't seem elegant, because I'm putting boilerplate code into many derived classes C, D, E that are siblings of B (not shown) that just calls the default copy constructor of each. Isn't that what the default copy constructor is for, if only it could be virtual?

有更好的方法吗？

创建虚拟赋值运算符是错误的，因为我不想让C赋值给B，B到D等，只需克隆B，C，D或E。

Making a virtual assignment operator would be wrong, as I don't want C to assign to B, B to D, etc, just clone B, C, D or E.

#include <iostream>
using namespace std;

class A {
public:
    virtual ~A() {}
    virtual A* clone()=0;
};

class B : public A {
    int i;
    public:
    virtual A* clone() { 
        cout << "cloned B" << endl;
        return new B(*this);
    }
    virtual ~B() { cout << "destroyed b" << endl; }
};

int main() { 
    A* a = new B(); 
    A* aa = a->clone();
    delete a; 
    delete aa; 
    return 0;
}

推荐答案

克隆逻辑到层次结构中间的自己的类中：

You could always stick all the cloning logic into its own class in the middle of the hierarchy:

template <class Derived, class Base>
class CloneCRTP : public Base {
public:
    Derived* clone() const override {
        return new Derived(static_cast<Derived const&>(*this));
    }
};

然后：

class B : public CloneCRTP<B, A>
{
    int i;
public:
    virtual ~B() { cout << "destroyed b" << endl; }        
};

没有更多的样板。

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