关于c ++：如何从基类指针复制一个继承的类而不进行拼接？

how to copy an inherited class from its base class pointer without splicing?

我在尝试创建一个从指针到它的基类的类副本时遇到了一些困难。最好通过这个例子来说明：

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#include <iostream>
#include <vector>

class Base {
public:
Base() { }
virtual void test() { std::cout <<"I am just the base class
"; }
};

class First : public Base {
public:
First() { }
void test() { std::cout <<"This is the First class
"; }
};

class Second : public Base {
public:
Second() { }
void test() { std::cout <<"This is the Second class
"; }
};

int main() {
First *f = new First();
Second *s = new Second();

// First, I keep a vector of pointers to their base class
std::vector<Base *> ptrs;
ptrs.push_back(f);
ptrs.push_back(s);
ptrs[0]->test(); // Properly calls the implemented virtual class by inheritor
ptrs[1]->test(); // Properly calls the implemented virtual class by inheritor

// Now, I want to *copy* the class without it being spliced to just the base class
// from the ptrs vector (not from 'f')
First *f2 = new First(*ptrs[0]);
}

我最后得到的错误是：

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test.cpp: In function ‘int main()’:
test.cpp:35: error: no matching function for call to ‘First::First(Base&)’
test.cpp:12: note: candidates are: First::First()
test.cpp:10: note: First::First(const First&)

是否有任何方法可以强制转换此指针来复制完整的对象，而不仅仅是基类？或者需要存储一个指向继承者的指针才能使其工作？

您可以这样做：

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First *f2 = 0;
if ( typeid(*ptrs[0]) == typeid(First))
f2 = new First(*dynamic_cast<First*>(ptrs[0]));

那应该管用。

但更好的方法是在基类中使用clone()虚拟函数，并将其实现为派生类：

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class Base
{
public:
virtual ~Base() { } //virtual destructed added by me!
virtual Base *clone() = 0;
};

class First : public Base
{
public:
virtual First *clone() { /* implement it */ } //Covariant return type
};

和

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First *f2 = 0;
if ( typeid(*ptrs[0]) == typeid(First))
f2 = ptrs[0]->clone(); //no need to use new

需要注意的两点：

我已经向基类添加了虚拟析构函数。查看此主题以了解您可能需要它的原因。
我在派生类中对clone()使用了不同的返回类型。它被称为协变返回类型。

1 2	First fx=(First)ptrs[0]; First f2 = new First(fx);