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我有一个实现 原型模式的类​​层次结构，我想使用移动语义来限制对象的深度复制。我尝试使用move()成员函数来调整模式，这意味着我不再需要原始对象。这是我到目前为止所拥有的：

#include <iostream>
#include <utility>
#include <vector>

struct base
{
    virtual ~base() { }
    virtual base* clone() const = 0;
    virtual base* move() = 0;
};

struct derived1 : public base
{
    derived1()                { std::cout << "derived1::derived1()\n"; }
    derived1(const derived1&) { std::cout << "derived1::derived1(const derived1&)\n"; }
    derived1(derived1&&)      { std::cout << "derived1::derived1(derived1&&)\n"; }

    virtual ~derived1() { }

    virtual base* clone() const { return new derived1(*this); }
    virtual base* move()        { return new derived1(std::move(*this)); }
};

struct derived2 : public base
{
    derived2()                { std::cout << "derived2::derived2()\n"; }
    derived2(const derived2&) { std::cout << "derived2::derived2(const derived2&)\n"; }
    derived2(derived2&&)      { std::cout << "derived2::derived2(derived2&&)\n"; }

    virtual ~derived2() { }

    virtual base* clone() const { return new derived2(*this); }
    virtual base* move()        { return new derived2(std::move(*this)); }
};

std::vector<base*> vec;

void foo(const base& obj)
{
    vec.push_back(obj.clone());
}

void foo(base&& obj)
{
    vec.push_back(obj.move());
}

int main()
{
    derived1 d1;
    derived2 d2;

    foo(d1);
    foo(d2);
    foo(derived1());
    foo(derived2());
}

当我运行它时，它表明使用了好的构造函数：

derived1::derived1()
derived2::derived2()
derived1::derived1(const derived1&)
derived2::derived2(const derived2&)
derived1::derived1()
derived1::derived1(derived1&&)
derived2::derived2()
derived2::derived2(derived2&&)

到目前为止，似乎还不错。我只是不确定这是否是右值引用的标准兼容用法。有没有一点我没有想到会产生不良结果？

4

1 回答 1

1

对于重复的方法定义，我更喜欢CRTP。对于您的情况，我会声明如下：

template<typename TDerived>
class virtually_clonable : public base
{
public:
    virtual base* clone() override
    {
        return new TDerived(*AsDerived());
    }

    virtual base* move() override
    {
        return new TDerived(std::move(*AsDerived()));
    }
private:
    TDerived* AsDerived()
    {
        return static_cast<TDerived*>(this);
    }
};

在实现类时：

class derived1 : public virtually_clonable<derived1>
{
public:
/* your ctors goes here*/
/* no need to implement clone/move again */
};

class derived2 : public virtually_clonable<derived2>
{
public:
/* your ctors goes here*/
/* no need to implement clone/move again */
};

顺便说一句，您可能希望返回shared_ptr对象而不是原始指针。这通常是可克隆类型而不是unique_ptr.

于 2013-01-31T23:39:56.730 回答