使std :: vector分配对齐的内存

Question

是否可以使用std::vector自定义结构分配对齐的内存以便使用SIMD指令进行进一步处理？如果有可能,有Allocator没有人碰巧有这样的分配器,他可以分享？

Answer 1

编辑:我删除了std::allocatorGManNickG建议的继承,并使对齐参数成为编译时间.

我最近写了这段代码.它没有像我希望的那样进行测试,所以继续报告错误.:-)

enum class Alignment : size_t
{
    Normal = sizeof(void*),
    SSE    = 16,
    AVX    = 32,
};


namespace detail {
    void* allocate_aligned_memory(size_t align, size_t size);
    void deallocate_aligned_memory(void* ptr) noexcept;
}


template <typename T, Alignment Align = Alignment::AVX>
class AlignedAllocator;


template <Alignment Align>
class AlignedAllocator<void, Align>
{
public:
    typedef void*             pointer;
    typedef const void*       const_pointer;
    typedef void              value_type;

    template <class U> struct rebind { typedef AlignedAllocator<U, Align> other; };
};


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

    typedef std::true_type propagate_on_container_move_assignment;

    template <class U>
    struct rebind { typedef AlignedAllocator<U, Align> other; };

public:
    AlignedAllocator() noexcept
    {}

    template <class U>
    AlignedAllocator(const AlignedAllocator<U, Align>&) noexcept
    {}

    size_type
    max_size() const noexcept
    { return (size_type(~0) - size_type(Align)) / sizeof(T); }

    pointer
    address(reference x) const noexcept
    { return std::addressof(x); }

    const_pointer
    address(const_reference x) const noexcept
    { return std::addressof(x); }

    pointer
    allocate(size_type n, typename AlignedAllocator<void, Align>::const_pointer = 0)
    {
        const size_type alignment = static_cast<size_type>( Align );
        void* ptr = detail::allocate_aligned_memory(alignment , n * sizeof(T));
        if (ptr == nullptr) {
            throw std::bad_alloc();
        }

        return reinterpret_cast<pointer>(ptr);
    }

    void
    deallocate(pointer p, size_type) noexcept
    { return detail::deallocate_aligned_memory(p); }

    template <class U, class ...Args>
    void
    construct(U* p, Args&&... args)
    { ::new(reinterpret_cast<void*>(p)) U(std::forward<Args>(args)...); }

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


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

    typedef std::true_type propagate_on_container_move_assignment;

    template <class U>
    struct rebind { typedef AlignedAllocator<U, Align> other; };

public:
    AlignedAllocator() noexcept
    {}

    template <class U>
    AlignedAllocator(const AlignedAllocator<U, Align>&) noexcept
    {}

    size_type
    max_size() const noexcept
    { return (size_type(~0) - size_type(Align)) / sizeof(T); }

    const_pointer
    address(const_reference x) const noexcept
    { return std::addressof(x); }

    pointer
    allocate(size_type n, typename AlignedAllocator<void, Align>::const_pointer = 0)
    {
        const size_type alignment = static_cast<size_type>( Align );
        void* ptr = detail::allocate_aligned_memory(alignment , n * sizeof(T));
        if (ptr == nullptr) {
            throw std::bad_alloc();
        }

        return reinterpret_cast<pointer>(ptr);
    }

    void
    deallocate(pointer p, size_type) noexcept
    { return detail::deallocate_aligned_memory(p); }

    template <class U, class ...Args>
    void
    construct(U* p, Args&&... args)
    { ::new(reinterpret_cast<void*>(p)) U(std::forward<Args>(args)...); }

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

template <typename T, Alignment TAlign, typename U, Alignment UAlign>
inline
bool
operator== (const AlignedAllocator<T,TAlign>&, const AlignedAllocator<U, UAlign>&) noexcept
{ return TAlign == UAlign; }

template <typename T, Alignment TAlign, typename U, Alignment UAlign>
inline
bool
operator!= (const AlignedAllocator<T,TAlign>&, const AlignedAllocator<U, UAlign>&) noexcept
{ return TAlign != UAlign; }

实际分配调用的实现只是posix,但您可以轻松扩展.

void*
detail::allocate_aligned_memory(size_t align, size_t size)
{
    assert(align >= sizeof(void*));
    assert(nail::is_power_of_two(align));

    if (size == 0) {
        return nullptr;
    }

    void* ptr = nullptr;
    int rc = posix_memalign(&ptr, align, size);

    if (rc != 0) {
        return nullptr;
    }

    return ptr;
}


void
detail::deallocate_aligned_memory(void *ptr) noexcept
{
    return free(ptr);
}

需要C++ 11,顺便说一下.

Answer 2

在即将发布的1.56版本中,Boost库将包含Boost.Align.它提供的其他内存对齐帮助程序boost::alignment::aligned_allocator,可以使用直接替换,std::allocator并允许您指定对齐.请参阅https://boostorg.github.io/align/上的文档

Answer 3

从 C++17 开始，只需将std::vector<__m256i>or 与任何其他对齐类型一起使用。有 的对齐版本operator new，它用于std::allocator对齐类型（以及普通new表达式，因此 new__m256i[N]在 C++17 中启动也是安全的）。

@MarcGlisse 有一条评论这么说，使其成为一个答案，使其更加明显。