c++ - 使用公共(public)命名方法实现非公共(public)赋值运算符？

Question

它应该复制一个 AnimatedSprite。我重新考虑它具有更改 *this 对象的不幸副作用。

如何在没有副作用的情况下实现此功能？

编辑:

根据新的答案，问题实际上应该是:如何使用公共(public)命名方法 无副作用 实现非公共(public)赋值运算符？ (更改标题)。

public:
AnimatedSprite& AnimatedSprite::Clone(const AnimatedSprite& animatedSprite) {
    return (*this = animatedSprite);
}

protected:
AnimatedSprite& AnimatedSprite::operator=(const AnimatedSprite& rhs) {
    if(this == &rhs) return *this;

    destroy_bitmap(this->_frameImage);
    this->_frameImage = create_bitmap(rhs._frameImage->w, rhs._frameImage->h);
    clear_bitmap(this->_frameImage);
    this->_frameDimensions = rhs._frameDimensions;
    this->CalcCenterFrame();
    this->_frameRate = rhs._frameRate;
    if(rhs._animation != nullptr) {
        delete this->_animation;
        this->_animation = new a2de::AnimationHandler(*rhs._animation);
    } else {
        delete this->_animation;
        this->_animation = nullptr;
    }

    return *this;
}

Answer 1

您可以调用私有(private)分配运算符(operator):

public:
AnimatedSprite& AnimatedSprite::Clone(const AnimatedSprite& animatedSprite) {
    return ( operator=(animatedSprite));
}

无法绕过修改 this如果你正在尝试做作业

<罢工>通常，clone 返回指向新实例的指针或智能指针:

struct IFoo {
  virtual IFoo* clone() const = 0;
};
struct Foo1 : public virtual IFoo {
  virtual IFoo* clone() { return new Foo1(this);}
};
struct Foo2 : public virtual IFoo {
  virtual IFoo* clone() { return new Foo2(this);}
};

IFoo* foo0 = new Foo1();
...
IFoo* fooClone = foo0.clone();

<罢工>