派生的 C++ 如何能?类通过基指针克隆自身?

发布于 2024-09-07 13:37:36 字数 483 浏览 3 评论 0原文

这就是我想要做的(此代码不起作用):

class Base
{
    virtual Base *clone() { return new Base(this); }
    virtual void ID() { printf("BASE");
};

class Derived : publc Base
{
    virtual Base *clone() { return new Derived(this); }
    virtual void ID() { printf("DERIVED"); }
}

.
.
Derived d;
Base *bp = &d;
Base *bp2 = bp->clone();

bp2->ID();

喜欢是看到“DERIVED”打印出来......我得到的是“BASE”。我是一名长期的 C 程序员,并且对 C++ 相当有经验...但我在这方面没有取得任何进展...任何帮助将不胜感激。

Here's what I'm trying to do (this code doesn't work):

class Base
{
    virtual Base *clone() { return new Base(this); }
    virtual void ID() { printf("BASE");
};

class Derived : publc Base
{
    virtual Base *clone() { return new Derived(this); }
    virtual void ID() { printf("DERIVED"); }
}

.
.
Derived d;
Base *bp = &d;
Base *bp2 = bp->clone();

bp2->ID();

What I'd like is to see "DERIVED" printed out... what I get is "BASE". I'm a long-time C programmer, and fairly experienced with C++... but I'm not making any headway with this one... any help would be appreciated.

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评论(6

长发绾君心 2024-09-14 13:37:36

一旦所有的编译错误都被修复,我最终得到了这个:

#include <cstdio>

class Base
{
  public:
    Base() {}
    Base(const Base&) {}
    virtual Base *clone() { return new Base(*this); }
    virtual void ID() { printf("BASE"); }
};

class Derived : public Base
{
  public:
    Derived() {}
    Derived(const Derived&) {}
    virtual Base *clone() { return new Derived(*this); }
    virtual void ID() { printf("DERIVED"); }
};


int main()
{
  Derived d;
  Base *bp = &d;
  Base *bp2 = bp->clone();

  bp2->ID();
}

这给了你你正在寻找的东西——派生的。

Once all the compile errors are fixed, I ended up with this:

#include <cstdio>

class Base
{
  public:
    Base() {}
    Base(const Base&) {}
    virtual Base *clone() { return new Base(*this); }
    virtual void ID() { printf("BASE"); }
};

class Derived : public Base
{
  public:
    Derived() {}
    Derived(const Derived&) {}
    virtual Base *clone() { return new Derived(*this); }
    virtual void ID() { printf("DERIVED"); }
};


int main()
{
  Derived d;
  Base *bp = &d;
  Base *bp2 = bp->clone();

  bp2->ID();
}

Which gives you what you are looking for -- DERIVED.

木格 2024-09-14 13:37:36

该代码充满了语法错误。也许最重要的是,Derived 并不继承自 Base。其次,除了语法错误(可能是简单的拼写错误)之外,Base 显然需要一个虚拟析构函数。克隆方法几乎要求您可以在基指针(Base*)上调用删除操作符。

class Base
{
public:
    virtual ~Base() {}
    virtual Base* clone() const { return new Base(*this); }
    virtual void ID() const { printf("BASE"); }
};

class Derived: public Base
{
public:
    // [Edit] Changed return type to Derived* instead of Base*.
    // Thanks to Matthieu for pointing this out. @see comments below.
    virtual Derived* clone() const { return new Derived(*this); }
    virtual void ID() const { printf("DERIVED"); }
};

int main()
{
    Derived d;
    Base* bp = &d;

    Base* bp2 = bp->clone();
    bp2->ID(); // outputs DERIVED as expected
    delete bp2;
}

That code is riddled with syntactical errors. Perhaps most significantly, Derived doesn't inherit from Base. Secondly, aside from the syntactical errors (probably simple typos), Base obviously needs a virtual destructor. The clone method pretty much demands that you can call operator delete on a base pointer (Base*).

class Base
{
public:
    virtual ~Base() {}
    virtual Base* clone() const { return new Base(*this); }
    virtual void ID() const { printf("BASE"); }
};

class Derived: public Base
{
public:
    // [Edit] Changed return type to Derived* instead of Base*.
    // Thanks to Matthieu for pointing this out. @see comments below.
    virtual Derived* clone() const { return new Derived(*this); }
    virtual void ID() const { printf("DERIVED"); }
};

int main()
{
    Derived d;
    Base* bp = &d;

    Base* bp2 = bp->clone();
    bp2->ID(); // outputs DERIVED as expected
    delete bp2;
}
豆芽 2024-09-14 13:37:36

使用 Base bp = &d;

您已经“切片”了 d,因此对于编译器来说,bp 实际上只是 Base,这就是为什么当您调用 bp->clone() 时,编译器会调用 Base::clone();bp2- >ID() 打印 BASE

基础& bp = d; 会做你想做的事。

With Base bp = &d;

You've "sliced" d, so to the compiler, bp really is only of type Base, which is why when you call bp->clone() the compiler calls Base::clone(); and bp2->ID() prints BASE.

Base& bp = d; will do what you want.

苏别ゝ 2024-09-14 13:37:36

您正在切片Base bp = &d; (这会从导出的指针构造一个新的基本 bp。)

尝试使用 Base* bp = &d; 代替。 (即创建一个指向 Derived 对象的 Base 类型的指针。)

You're slicing the class in Base bp = &d; (this constructs a new base bp from the derived-ptr.)

Try Base* bp = &d; instead. (i.e. create a pointer of Base type to the Derived object.)

巾帼英雄 2024-09-14 13:37:36

您的示例不正确,无法编译。具体来说这一行:

Base bp = &d;

这也可能是您问题的根本原因(您可能正在切片您的对象),但如果没有看到工作代码,我无法确定。

您还遇到一个问题,即您的两个类不相关(您的意思是编写 class Derived : public Base 吗?)

Your example is incorrect and will not compile. Specifically this line:

Base bp = &d;

That may also be the root cause of your problem (you may be slicing your object), but I can't tell for certain without seeing working code.

You also have a problem where your two classes are not related (did you mean to write class Derived : public Base?)

彩扇题诗 2024-09-14 13:37:36

除了愚蠢的语法拼写错误和缺失的因素之外,代码看起来不错。

The code looks fine, other than the silly syntax typos and the missing ctors.

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