C++:严格别名与联合滥用

Question

提前道歉,这可能是一个蹩脚的第一篇文章.虽然这个主题有很多材料,但对我来说很少是明确的和/或可理解的.

我有一个AlignedArray模板类,可以在堆上动态分配任意对齐的内存(我需要为AVX程序集例程进行32字节对齐).这需要一些丑陋的指针操作.

Agner Fog在cppexamples.zip中提供了一个样本类,它滥用了一个联合(http://www.agner.org/optimize/optimization_manuals.zip).但是,我知道写一个联盟的一个成员然后从另一个成员读取会导致UB.

AFAICT可以安全地将任何指针类型别名为a char *,但只能在一个方向上.这是我理解变得模糊的地方.这是我AlignedArray 课程的精简版(基本上是对Agner的重写,以帮助我理解):

template <typename T, size_t alignment = 32>
class AlignedArray
{
    size_t m_size;
    char * m_unaligned;
    T * m_aligned;

public:
    AlignedArray (size_t const size)
        : m_size(0)
        , m_unaligned(0)
        , m_aligned(0)
    {
        this->size(size);
    }

    ~AlignedArray ()
    {
        this->size(0);
    }

    T const & operator [] (size_t const i) const { return m_aligned[i]; }

    T & operator [] (size_t const i) { return m_aligned[i]; }

    size_t const size () { return m_size; }

    void size (size_t const size)
    {
        if (size > 0)
        {
            if (size != m_size)
            {
                char * unaligned = 0;
                unaligned = new char [size * sizeof(T) + alignment - 1];
                if (unaligned)
                {
                    // Agner:
                    /*
                    union {
                        char * c;
                        T * t;
                        size_t s;
                    } aligned;
                    aligned.c = unaligned + alignment - 1;
                    aligned.s &= ~(alignment - 1);
                    */

                    // Me:
                    T * aligned = reinterpret_cast<T *>((reinterpret_cast<size_t>(unaligned) + alignment - 1) & ~(alignment - 1));

                    if (m_unaligned)
                    {
                        // Agner:
                        //memcpy(aligned.c, m_aligned, std::min(size, m_size));

                        // Me:
                        memcpy(aligned, m_aligned, std::min(size, m_size));

                        delete [] m_unaligned;
                    }
                    m_size = size;
                    m_unaligned = unaligned;

                    // Agner:
                    //m_aligned = aligned.t;

                    // Me:
                    m_aligned = aligned;
                }
                return;
            }
            return;
        }
        if (m_unaligned)
        {
            delete [] m_unaligned;
            m_size = 0;
            m_unaligned = 0;
            m_aligned = 0;
        }
    }
};

那么哪种方法是安全的(r)？

Answer 1

我有实现（替换）new和delete运算符的代码，适用于 SIMD（即 SSE / AVX）。它使用以下您可能会觉得有用的函数：

static inline void *G0__SIMD_malloc (size_t size)
{
    constexpr size_t align = G0_SIMD_ALIGN;
    void *ptr, *uptr;

    static_assert(G0_SIMD_ALIGN >= sizeof(void *),
                  "insufficient alignment for pointer storage");

    static_assert((G0_SIMD_ALIGN & (G0_SIMD_ALIGN - 1)) == 0,
                  "G0_SIMD_ALIGN value must be a power of (2)");

    size += align; // raw pointer storage with alignment padding.

    if ((uptr = malloc(size)) == nullptr)
        return nullptr;

    // size_t addr = reinterpret_cast<size_t>(uptr);
    uintptr_t addr = reinterpret_cast<uintptr_t>(uptr);

    ptr = reinterpret_cast<void *>
        ((addr + align) & ~(align - 1));

    *(reinterpret_cast<void **>(ptr) - 1) = uptr; // (raw ptr)

    return ptr;
}


static inline void G0__SIMD_free (void *ptr)
{
    if (ptr != nullptr)
        free(*(reinterpret_cast<void **>(ptr) - 1)); // (raw ptr)
}

这应该很容易适应。显然，您将替换mallocand free，因为您正在使用全局newanddelete进行原始（字符）存储。它假设size_t对于地址算术来说足够宽——实际上是正确的，但uintptr_tfrom<cstdint>会更正确。