Question

有没有办法确保valarray使用对齐的内存，以便可以使用SSE和AVX进行矢量化？据我所知，STL不保证对齐，你可以不将分配器传递给valarray。还有另一种方法可以达到这个目的吗？

提前致谢！

Answer 1

我通常将std::vector与我自己的分配器一起使用，分配器具有对齐作为模板参数，并调用_mm_malloc()或_aligned_malloc()。这非常有效，也适用于AVX（32字节对齐）。适当编写的模板化用户代码会自动选择所需的对齐方式。

在AlignmentAllocator<>和帮助程序的代码下面。在gcc和icpc下测试。

/// allocate and de-allocate aligned memory
template<std::size_t alignment>
struct static_allocator {
  static void*allocate(std::size_t n)
  {
    if(n == 0) return 0;
    if(n > max_size())
      throw std::bad_alloc();
    void*ret =
#if defined(__GNUC__) || defined (__INTEL_COMPILER)
      _mm_malloc
#else
      _aligned_malloc
#endif
      (n,alignment);
    if(!ret)
      throw std::bad_alloc();
    return ret;
  }
  static void deallocate(void*p)
  {
#if defined(__GNUC__) || defined (__INTEL_COMPILER)
    _mm_free
#else
    _aligned_free
#endif
    (p);
  }
  static std::size_t max_size ()
  { return std::numeric_limits<std::size_t>::max(); }
};

/// allocate and de-allocate unaligned memory
template<>
struct static_allocator<1> {
  static std::size_t max_size () noexcept
  { return std::numeric_limits<std::size_t>::max(); }
  static void*allocate(std::size_t n)
  { 
    if(n == 0) return 0;
    void*ret = new char[n];
    return ret;
  }
  static void deallocate(void*p)
  { delete[] static_cast<char*>(p); }
};

template<> struct static_allocator<0>;

/// allocator with explicit alignment
template<typename _Tp, std::size_t alignment = 16>
class AlignmentAllocator
{
  typedef static_allocator<alignment> static_alloc;
public:
  typedef size_t     size_type;
  typedef ptrdiff_t  difference_type;
  typedef _Tp*       pointer;
  typedef const _Tp* const_pointer;
  typedef _Tp&       reference;
  typedef const _Tp& const_reference;
  typedef _Tp        value_type;

  template <typename _Tp1>
  struct rebind
  { typedef AlignmentAllocator<_Tp1, alignment> other; };

  AlignmentAllocator() {}

  AlignmentAllocator(const AlignmentAllocator&) {}

  template <typename _Tp1>
  AlignmentAllocator(const AlignmentAllocator<_Tp1, alignment> &) {}

  ~AlignmentAllocator() {}

  pointer address (reference x) const
  {
#if __cplusplus >= 201103L
    return std::addressof(x);
#else
    return reinterpret_cast<_Tp*>(&reinterpret_cast<char&>(x));
#endif
  }

  const_pointer address (const_reference x) const
  {
#if __cplusplus >= 201103L
    return std::addressof(x);
#else
    return reinterpret_cast<const _Tp*>(&reinterpret_cast<const char&>(x));
#endif
  }

  pointer allocate (size_type n, const void* = 0)
  { return static_cast<pointer>(static_alloc::allocate(n*sizeof(value_type))); }

  void deallocate (pointer p, size_type)
  { static_alloc::deallocate(p); }

  size_type max_size () const
  { return static_alloc::max_size() / sizeof (value_type); }

#if __cplusplus >= 201103L

  template<typename _Up, typename... _Args>
  void construct(_Up* p, _Args&&... args)
  { ::new(static_cast<void*>(p)) _Up(std::forward<_Args>(args)...); }

  template<typename _Up>
  void destroy(_Up* p)
  { p->~_Up(); }

#else

  void construct (pointer p, const_reference val)
  { ::new(static_cast<void*>(p)) value_type(val); }

  void destroy (pointer p)
  { p->~value_type (); }

#endif

  bool operator!=(const AlignmentAllocator&) const 
  { return false; }

  // Returns true if and only if storage allocated from *this
  // can be deallocated from other, and vice versa.
  // Always returns true for stateless allocators.
  bool operator==(const AlignmentAllocator&) const 
  { return true; }

};// class AlignmentAllocator<>

/// AlignmentAllocator<void> specialization.
template<std::size_t alignment>
class AlignmentAllocator<void, alignment>
{
public:
  typedef size_t      size_type;
  typedef ptrdiff_t   difference_type;
  typedef void*       pointer;
  typedef const void* const_pointer;
  typedef void        value_type;

  template<typename _Tp1>
  struct rebind
  { typedef AlignmentAllocator<_Tp1, alignment> other; };
};

用于SSE / AVX的对齐内存上的valarray

1 个答案: