Question

如果要在C ++中实现Clone模式，他可能不确定安全性，因为派生类可能会忘记覆盖它：

struct A {
    virtual A* Clone() const {
        return new A(*this);
    }
}

struct B : A {
    int value;
};

int main() {
   B b;
   // oops
   auto b_clone = b.Clone();
   delete b_clone;
}

在这方面，可以使用哪些方法来改善C ++中的Clone模式？

提出了一个更一般的问题： Forcing a derived class to overload a virtual method in a non-abstract base class

但是，在C ++中找到好的解决方案似乎太笼统了-讨论是关于强制方法重写的可能方法。我对发现有用的模式更感兴趣，这可能对使用Cloneable模式的确切情况有所帮助。

Answer 1

这是对答案之一的阐述，建议使用typeid进行运行时检查： Forcing a derived class to overload a virtual method in a non-abstract base class

使用CRTP，可以提出以下基本思想：

创建类Cloneable<Derived>，该类管理Derived的克隆，并添加所有必需的运行时检查（似乎即使使用CRTP也无法进行编译时检查）。

但是，这并非不重要，而且还必须通过Cloneable管理继承，如下所述：

#include <memory>
#include <cassert>
#include <type_traits>
#include <typeinfo>

class CloneableInterface {
public:
    virtual std::unique_ptr<CloneableInterface> Clone() const = 0;
};

template <class... inherit_from>
struct InheritFrom : public inherit_from... {
};

template <class Derived, class AnotherBase = void, bool base_is_cloneable = std::is_base_of_v<CloneableInterface, AnotherBase>>
class Cloneable;

// three identical implementations, only the inheritance is different

// "no base is defined" case
template <class Derived>
class Cloneable<Derived, void, false> : public CloneableInterface {
public:
    std::unique_ptr<CloneableInterface> Clone() const override {
        assert(typeid(*this) == typeid(Derived));
    return std::make_unique<Derived>(static_cast<const Derived&>(*this));
    }
};

// Base is defined, and already provides CloneableInterface
template <class Derived, class AnotherBase>
class Cloneable<Derived, AnotherBase, true> : public AnotherBase {
   ...
};

// Base is defined, but has no CloneableInterface
template <class Derived, class AnotherBase>
class Cloneable<Derived, AnotherBase, false> : public AnotherBase, public CloneableInterface {
    ...
};

用法示例：

class Base : public Cloneable<Base> {
};

// Just some struct to test multiple inheritance
struct Other {
};

struct Derived : Cloneable<Derived, InheritFrom<Base, Other>> {
};

struct OtherBase {
};

struct OtherDerived : Cloneable<OtherDerived, InheritFrom<OtherBase>> {
};

int main() {
    // compiles and runs
    auto base_ptr = std::make_unique<Base>();
    auto derived_ptr = std::make_unique<Derived>();
    auto base_clone = base_ptr->Clone();
    auto derived_clone = derived_ptr->Clone();

    auto otherderived_ptr = std::make_unique<OtherDerived>();
    auto otherderived_clone = otherderived_ptr->Clone();
}

欢迎任何批评家和改进建议！

Answer 2

C ++ 17及更高版本提供了std::any。从理论上讲，您可以制作一个克隆函数，而不是返回std::any*：

struct A {
    virtual std::any* Clone() const {
        return new A(*this);
    }
}

struct B : A {
    int value;
    // I suppose it doesn't have to be virtual here, 
    // but just in case we want to inherit the cloning capability from B as well
    virtual std::any* Clone() const { // Note: you still need to override this function
        return new B(*this);          // in the lower levels, though
    }
};
// Note: I'm still on MSVS2010, so this C++17 code is untested.
// Particularly problematic could be this main
int main() {
   B b;
   // Here is the clone
   auto b_clone = std::any_cast<B*>(b.Clone());
   delete b_clone;
}

再次，这是未经测试的，但在理论上应该可行。

提高克隆模式的安全性

2 个答案: