避免标准容器中元素的默认构造

Question

我有兴趣构建一个uninitialized_vector容器,它在语义上std::vector与警告相同,否则将使用无参数构造函数创建的新元素将在没有初始化的情况下创建.我主要是想避免将POD初始化为0. 据我所知,通过std::vector与一种特殊的分配器结合,无法实现这一点.

我想以同样的方式构建我的容器std::stack,它适应用户提供的容器(在我的情况下std::vector).换句话说,我想避免重新实现整体,std::vector而是围绕它提供一个"立面".

有没有一种简单的方法来控制"外部"的默认构造std::vector？

---

这是我到达的解决方案,这启发了Kerrek的答案:

#include <iostream>
#include <vector>
#include <memory>
#include <algorithm>
#include <cassert>

// uninitialized_allocator adapts a given base allocator
// uninitialized_allocator's behavior is equivalent to the base
// except for its no-argument construct function, which is a no-op
template<typename T, typename BaseAllocator = std::allocator<T>>
  struct uninitialized_allocator
    : BaseAllocator::template rebind<T>::other
{
  typedef typename BaseAllocator::template rebind<T>::other super_t;

  template<typename U>
    struct rebind
  {
    typedef uninitialized_allocator<U, BaseAllocator> other;
  };

  // XXX for testing purposes
  typename super_t::pointer allocate(typename super_t::size_type n)
  {
    auto result = super_t::allocate(n);

    // fill result with 13 so we can check afterwards that
    // the result was not default-constructed
    std::fill(result, result + n, 13);
    return result;
  }

  // catch default-construction
  void construct(T *p)
  {
    // no-op
  }

  // forward everything else with at least one argument to the base
  template<typename Arg1, typename... Args>
    void construct(T* p, Arg1 &&arg1, Args&&... args)
  {
    super_t::construct(p, std::forward<Arg1>(arg1), std::forward<Args>(args)...);
  }
};

namespace std
{

// XXX specialize allocator_traits
//     this shouldn't be necessary, but clang++ 2.7 + libc++ has trouble
//     recognizing that uninitialized_allocator<T> has a well-formed
//     construct function
template<typename T>
  struct allocator_traits<uninitialized_allocator<T> >
    : std::allocator_traits<std::allocator<T>>
{
  typedef uninitialized_allocator<T> allocator_type;

  // for testing purposes, forward allocate through
  static typename allocator_type::pointer allocate(allocator_type &a, typename allocator_type::size_type n)
  {
    return a.allocate(n);
  }

  template<typename... Args>
    static void construct(allocator_type &a, T* ptr, Args&&... args)
  {
    a.construct(ptr, std::forward<Args>(args)...);
  };
};

}

// uninitialized_vector is implemented by adapting an allocator and
// inheriting from std::vector
// a template alias would be another possiblity

// XXX does not compile with clang++ 2.9
//template<typename T, typename BaseAllocator>
//using uninitialized_vector = std::vector<T, uninitialized_allocator<T,BaseAllocator>>;

template<typename T, typename BaseAllocator = std::allocator<T>>
  struct uninitialized_vector
    : std::vector<T, uninitialized_allocator<T,BaseAllocator>>
{};

int main()
{
  uninitialized_vector<int> vec;
  vec.resize(10);

  // everything should be 13
  assert(std::count(vec.begin(), vec.end(), 13) == vec.size());

  // copy construction should be preserved
  vec.push_back(7);
  assert(7 == vec.back());

  return 0;
}

该解决方案将根据特定供应商的编译器和STL的std::vector实现与c ++ 11的符合程度而有效.

Answer 1

不要在容器周围使用包装器,而是考虑在元素类型周围使用包装器:

template <typename T>
struct uninitialized
{
    uninitialized() { }
    T value;
};

Answer 2

我认为问题归结为容器对元素执行的初始化类型.相比:

T * p1 = new T;   // default-initalization
T * p2 = new T(); // value-initialization

标准容器的问题在于它们将默认参数初始化为值,如resize(size_t, T = T()).这意味着没有优雅的方法来避免值初始化或复制.(类似于构造函数.)

即使使用标准分配器也不起作用,因为它们的中心construct()函数采用了一个变为值初始化的参数.您更需要的是construct()使用默认初始化:

template <typename T>
void definit_construct(void * addr)
{
  new (addr) T;  // default-initialization
}

这样的东西不再是一个符合标准的分配器,但你可以围绕这个想法建立自己的容器.