为什么不从 std::allocator 继承

问题描述

我像这样创建了自己的分配器:

I created my own allocator like so:

template<typename T>
class BasicAllocator
{
    public:
        typedef size_t size_type;
        typedef ptrdiff_t difference_type;
        typedef T* pointer;
        typedef const T* const_pointer;
        typedef T& reference;
        typedef const T& const_reference;
        typedef T value_type;


        BasicAllocator() throw() {};
        BasicAllocator (const BasicAllocator& other) throw() {};

        template<typename U>
        BasicAllocator (const BasicAllocator<U>& other) throw() {};

        template<typename U>
        BasicAllocator& operator = (const BasicAllocator<U>& other) {return *this;}
        BasicAllocator<T>& operator = (const BasicAllocator& other) {return *this;}
        ~BasicAllocator() {}

        pointer address (reference value) const {return &value;}
        const_pointer address (const_reference value) const {return &value;}

        pointer allocate (size_type n, const void* hint = 0) {return static_cast<pointer> (::operator new (n * sizeof (value_type) ) );}
        void deallocate (void* ptr, size_type n) {::operator delete (static_cast<T*> (ptr) );}

        template<typename U, typename... Args>
        void construct (U* ptr, Args&&  ... args) {::new (static_cast<void*> (ptr) ) U (std::forward<Args> (args)...);}
        void construct (pointer ptr, const T& val) {new (static_cast<T*> (ptr) ) T (val);}

        template<typename U>
        void destroy (U* ptr) {ptr->~U();}
        void destroy (pointer ptr) {ptr->~T();}

        size_type max_size() const {return std::numeric_limits<std::size_t>::max() / sizeof (T);} /**return std::size_t(-1);**/

        template<typename U>
        struct rebind
        {
            typedef BasicAllocator<U> other;
        };
};

但我想知道为什么我永远不应该从 std::allocator 继承.是因为它没有虚拟析构函数吗?我看过很多帖子说应该创建自己的而不是继承.我明白为什么我们不应该继承 std::string 和 std::vector 但是继承 std::allocator 有什么问题?

But I want to know why I should never inherit from std::allocator. Is it because it doesn't have a virtual destructor? I've seen many posts saying that one should create their own and not inherit. I understand why we shouldn't inherit std::string and std::vector but what is wrong with inheriting std::allocator?

我可以继承上面的类吗?或者我需要一个虚拟析构函数来做到这一点?

Can I inherit my class above? Or do I need a virtual destructor to do that?

为什么?

推荐答案

很多人会在这个帖子里发帖说你不应该从 std::allocator 继承，因为它没有有一个虚拟析构函数.他们将讨论多态性以及通过指向基类的指针进行切片和删除，这些都不是标准第 17.6.3.5 节 [allocator.requirements] 中详述的分配器要求所允许的.除非有人证明从 std::allocator 派生的类不能满足其中一个要求，否则就是简单的货物崇拜心态.

A lot of people are going to post in this thread that you should not inherit from std::allocator because it doesn't have a virtual destructor. They'll talk about polymorphism and slicing and deleting via pointer-to-base class, none of which are permitted by the allocator requirements as detailed in section 17.6.3.5 [allocator.requirements] of the standard. Until someone demonstrates that a class derived from std::allocator fails to meet one of those requirements, it's simple cargo cult mentality.

也就是说，在 C++11 中几乎没有理由从 std::allocator 派生.C++11 对分配器的大修引入了特征模板 std::allocator_traits 位于分配器和它的用户之间，并通过模板元编程为许多必需的功能提供合理的默认值.C++11 中的最小分配器可以很简单:

That said, there is little reason to derive from std::allocator in C++11. C++11's overhaul of allocators introduced the traits template std::allocator_traits to sit between an allocator and its users and provide reasonable defaults for many of the required features via template metaprogramming. A minimal allocator in C++11 can be as simple as:

template <typename T>
struct mallocator {
  using value_type = T;

  mallocator() = default;
  template <class U>
  mallocator(const mallocator<U>&) {}

  T* allocate(std::size_t n) {
    std::cout << "allocate(" << n << ") = ";
    if (n <= std::numeric_limits<std::size_t>::max() / sizeof(T)) {
      if (auto ptr = std::malloc(n * sizeof(T))) {
        return static_cast<T*>(ptr);
      }
    }
    throw std::bad_alloc();
  }
  void deallocate(T* ptr, std::size_t n) {
    std::free(ptr);
  }
};

template <typename T, typename U>
inline bool operator == (const mallocator<T>&, const mallocator<U>&) {
  return true;
}

template <typename T, typename U>
inline bool operator != (const mallocator<T>& a, const mallocator<U>& b) {
  return !(a == b);
}

std::allocator_traits 的正确使用尚未完全出现在所有标准库中.例如，当使用 GCC 4.8.1 编译时，上面的示例分配器不能与 std::list 一起正常工作 - std::list 代码抱怨缺少成员，因为还没更新.

Proper use of std::allocator_traits isn't fully present in all standard libraries yet. For example, the sample allocator above doesn't work correctly with std::list when compiled with GCC 4.8.1 - the std::list code complains about missing members since it hasn't been updated yet.

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