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C ++ STL容器中的多态性

[英]Polymorphism in C++ STL containers

I was wondering if an elegant solution exists to this problem. 我想知道这个问题是否存在优雅的解决方案。

Suppose the following: 假设如下:

class Base {
  private:
    int data;
  protected:
    Base(int data) : data(data) { }
    int getData() const { return data; }
    virtual bool someOp() = 0;
}

class Derived1 : public Base {
  private:
    int data2;
  public:
    Derived1(int data, int data2) : Base(data), data2(data2) { }
    int getData2() const { return data2; }
    bool someOp() override { /* something */ }
}

class Derived2 : public Base {
  private:
    float data2;
  public:
    Derived1(int data, float data2) : Base(data), data2(data2) { }
    float getData2() const { return data2; }
    bool someOp() override { /* something */ }
}

And suppose that I have the total control over the hierarchy so I can assume that Base won't get extended, nor any DerivedX class. 并且假设我对层次结构具有完全控制权,因此我可以假设Base不会被扩展,也不会有任何DerivedX类。

Now I want to store them in a std::vector , if I want to use polymorphism I'm forced to store pointers otherwise object slicing will prevent me to store the additional derived properties. 现在我想将它们存储在std::vector ,如果我想使用多态,我会被迫存储指针,否则对象切片将阻止我存储其他派生属性。 So I'm basically forced to use a std::vector<unique_ptr<Base>> . 所以我基本上被迫使用std::vector<unique_ptr<Base>>

But let's assume that I need to store plenty of these objects and I don't want to waste for double heap allocation (the internal std::vector + the object itself) and that at the same time I can assume that: 但是我们假设我需要存储大量这些对象,我不想浪费双堆分配(内部std::vector +对象本身),同时我可以假设:

  • the class hierarchy is perfectly defined and won't be extended without knowing it 类层次结构是完美定义的,如果不知道就不会扩展
  • sizeof(DerivedX) won't be so larger than sizeof(Base) sizeof(DerivedX)不会比sizeof(Base)

So I'm wondering if there is an elegant way to keep polymorphism and avoid storing pointers. 所以我想知道是否有一种优雅的方法来保持多态性并避免存储指针。 I could think of some solutions like 我可以想到一些类似的解决方案

class Base {
  enum {
   DERIVED1,
   DERIVED2
  } type;

  int data;
  union {
    int derived1data;
    float derived2data;
  }

  bool someOp() {
    if (this->type == DERIVED1) ...
    else if ..
  }
}

But this is clearly not elegant. 但这显然不优雅。 I could also try to exploit the fact that object slicing shouldn't occur if sizeof(Derived) == sizeof(Base) by using a protected union in Base and accessing it from Derived and casting the address to elements in the std::vector to the desired type (according to an enum) but this sounds ugly too. 我还可以尝试利用如下事实:如果sizeof(Derived) == sizeof(Base)通过在Base使用受保护的联合并从Derived访问它并将地址转换为std::vector元素来实现到期望的类型(根据枚举),但这听起来也很难看。

Type erasure and the small buffer optimization. 键入擦除和小缓冲区优化。

You can type erase almost any property in C++, creating a custom interface that "knows" how to apply the property to a now-unknown type. 您可以在C ++中键入几乎任何属性,创建一个“知道”如何将属性应用于现在未知类型的自定义接口。

boost::any type erases down to copy, destroy, get typeid, and cast-back-to-exactly-matching-type. boost::any类型都会删除以复制,销毁,获取typeid和强制转换为精确匹配类型。 std::function type erases down to copy, invoke with a specific signature, destroy, and cast-back-to-identical-type (the last is rarely used). std::function类型擦除以复制,使用特定签名调用,销毁和反向转换为相同类型(最后很少使用)。

Free store based type erasure implementations get move semantics 'for free' by swapping around pointers. 基于免费存储的类型擦除实现通过交换指针来“免费”移动语义。

In your case, you'll want to create a "large enough" aligned storage in the type. 在您的情况下,您将需要在类型中创建“足够大”的对齐存储。 You'll want to type erase down to copy, get-as-reference-to-base, destroy and probably move (as you are storing internally). 您需要键入erase down to copy,get-as-reference-to-base,destroy并可能移动(因为您在内部存储)。

std::aligned_storage is intended for your task (you can pass in the alignment requirements of all the types you intend to store). std::aligned_storage适用于您的任务(您可以传递要存储的所有类型的对齐要求)。 Then in-place new the object. 然后就地新建对象。

Build a table of the operations you want to perform on the object via void* -- copy, move, destroy, and convert-to- base* . 通过void* - 复制,移动,销毁和转换为base*构建要在对象上执行的操作的表。

template<class Sig>using f = Sig*;

struct table {
  f<base*(void*)>             convert_to_base;
  f<base const*(void const*)> convert_to_const_base;
  f<void(void*,void const*)>  copy_construct;
  f<void(void*)>              destroy;
  f<void(void*,void*)>        move_construct;
};
template<class T>
table const* get_table() {
  static table t = {
    // convert to base:
    [](void*p)->base*{
      T*t=static_cast<T*>(p);
      return t;
    },
    // convert to const base:
    [](void const*p)->base const*{
      T const*t=static_cast<T const*>(p);
      return t;
    },
    // etc
  };
  return &t;
}

now store get_table<T>() in your type-erased instance (it is basically a virtual function table, manually implemented), and write your wrapping regular class to use the functions from the table to manipulate the aligned_storage<?...> . 现在将get_table<T>()存储在类型擦除的实例中(它基本上是一个虚函数表,手动实现),并编写包装常规类以使用table的函数来操作aligned_storage<?...>

Now, this can be done easier by using boost::variant , or via something like my some type that acts like a any without using heap storage. 现在,通过使用boost::variant ,或者像我的某些类型,就像没有使用堆存储那样的any类型,可以更容易地做到这一点。 The some link includes an implementation that compiles of the pseudo-virtual function table technique above. 一些链接包括编译上面的伪虚函数表技术的实现。 I probably used aligned storage wrong, however, so be careful. 我可能错误地使用了对齐存储,所以要小心。

You could use std::aligned_storage to wrap your classes. 您可以使用std :: aligned_storage来包装您的类。 Assuming that Derived2 is the largest: 假设Derived2是最大的:

class Storage
{
public:
  Storage(int data, int data2)
  {
    new (&data) Derived1(data, data2);
  }
  Storage(int data, float data2)
  {
    new (&data) Derived2(data, data2);
  }
  Base* getBase()
  {
    return reinterpret_cast<Base*>(&data);
  }
  ~Storage()
  {
    getBase()->Base::~Base();
  }
private:
  std::aligned_storage<sizeof(Derived2)> data;
};

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