C ++ 11的decltype是否不需要克隆？ [英] Does C++11's decltype make clone unnecessary?

问题描述

克隆范例用于创建派生类的副本，而不向下转换为基类类型。不幸的是， clone 必须在每个子类中实现（或与CRTP混用）。

The clone paradigm is used to make a copy of a derived class without casting down to the base class type. Unfortunately, clone must be implemented in each subclass (or with a mixin with CRTP).

C ++ 11的 decltype 使这不必要？

Is there any chance that C++11's decltype makes this unnecessary?

我不认为下面的代码实际上复制了原始，但只是指向它的引用。当我尝试使用 new decltype（* original）时，我得到一个错误：
错误：new不能应用于引用类型。

I don't think the code below actually copies original, but simply points a reference to it. When I tried to use new decltype(*original), I get an error: error: new cannot be applied to a reference type.

克隆仍然可以在C ++ 11中使用吗？或者有一些新的方法使用RTTI从基类指针复制派生类对象？

Is clone still the way to go in C++11? Or is there some new way to use RTTI to copy a derived class object from a base class pointer?

#include <iostream>

struct Base
{
  virtual void print()
  {
    std::cout << "Base" << std::endl;
  }
};

struct Derived : public Base
{
  int val;
  Derived() {val=0;}
  Derived(int val_param): val(val_param) {}
  virtual void print()
  {
    std::cout << "Derived " << val << std::endl;
  }
};

int main() {
  Base * original = new Derived(1);
  original->print();

  // copies by casting down to Base: you need to know the type of *original
  Base * unworking_copy = new Base(*original);
  unworking_copy->print();

  decltype(*original) on_stack = *original;
  on_stack.print();
  return 0;
}

推荐答案

decltype 是一个静态构造。像所有的C ++类型构造一样，它不能推导出对象的 runtime 类型。 decltype（* original）只是 Base& 。

decltype is a static construct. Like all C++ typing constructs, it cannot deduce the runtime type of an object. decltype(*original) is just Base&.

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