在C ++中创建未知派生类的实例 [英] create instance of unknown derived class in C++

问题描述

我们假设我有一个指向一些基类的指针，我想创建这个对象的派生类的一个新实例。我如何做到这一点？

let's say I have a pointer to some base class and I want to create a new instance of this object's derived class. How can I do this?

class Base
{
    // virtual
};

class Derived : Base
{
    // ...
};


void someFunction(Base *b)
{
    Base *newInstance = new Derived(); // but here I don't know how I can get the Derived class type from *b
}

void test()
{
    Derived *d = new Derived();
    someFunction(d);
}

推荐答案

克隆

Cloning

struct Base {
   virtual Base* clone() { return new Base(*this); }
};

struct Derived : Base {
   virtual Base* clone() { return new Derived(*this); }
};


void someFunction(Base* b) {
   Base* newInstance = b->clone();
}

int main() {
   Derived* d = new Derived();
   someFunction(d);
}

这是一个非常典型的模式。

This is a pretty typical pattern.

struct Base {
   virtual Base* create_blank() { return new Base; }
};

struct Derived : Base {
   virtual Base* create_blank() { return new Derived; }
};


void someFunction(Base* b) {
   Base* newInstance = b->create_blank();
}

int main() {
   Derived* d = new Derived();
   someFunction(d);
}

虽然我不认为这是一个典型的事情，它看起来像一个代码气味。您确定需要吗？

Though I don't think that this a typical thing to do; it looks to me like a bit of a code smell. Are you sure that you need it?

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