在arm neon中高效地重新排列和组合16个3位数字

我有一个由16个无符号数字组成的数组,其中每个数字都小于8(例如,它可以由3位表示)。这十六个数字加载到 uint8x16_t q寄存器。我需要对它们进行重组和组合,类似于以下伪代码:

void reshuffleCombine(uint8_t src[16], uint64_t* dst)
{
    uint64 d = 0;

    d |= uint64(src[0])  << 45;
    d |= uint64(src[4])  << 42;
    d |= uint64(src[8])  << 39;
    d |= uint64(src[12]) << 36;

    d |= uint64(src[1])  << 33;
    d |= uint64(src[5])  << 30;
    d |= uint64(src[9])  << 27;
    d |= uint64(src[13]) << 24;

    d |= uint64(src[2])  << 21;
    d |= uint64(src[6])  << 18;
    d |= uint64(src[10]) << 15;
    d |= uint64(src[14]) << 12;

    d |= uint64(src[3])  << 9;
    d |= uint64(src[7])  << 6;
    d |= uint64(src[11]) << 3;
    d |= uint64(src[15]) << 0;

    *dst = d;
}

void reshuffleCombineNeon(uint8x16_t src, uint64_t* dst)
{
    uint64x1_t res;
    // ??
    vst1_u64(dst, res);
}

我可以用1个vld和1个vtbl对它们进行重组,但是,整个操作是最后的步骤之一,不会重复很多次(例如,1个vld不能在多个重组组合之间共享),因此如果可能的话,最好使用vtrn/vzip,并且可能比vld+vtbl更有效。然而,问题的关键是:如何将这16个3位数字合并为一个48位值(存储在64位uint中)。此函数在算法结束时运行,16个3位数字来自neon,函数结果存储在内存中。

My neon version :

void reshuffleCombineNeon(uint8x16_t src, uint32_t* dst)
{
    static const uint8_t idx0[] = { 15, 7, 14, 6, 13, 5, 12, 4 };
    static const uint8_t idx1[] = { 11, 3, 10, 2, 9,  1, 8,  0 };

    uint8x8x2_t y;
    y.val[0] = vget_low_u8(src);
    y.val[1] = vget_high_u8(src);

    uint8x8_t vidx0 = vld1_u8(idx0);
    uint8x8_t vidx1 = vld1_u8(idx1);
    uint8x8_t x0 = vtbl2_u8(y, vidx0);
    uint8x8_t x1 = vtbl2_u8(y, vidx1);

    uint8x8_t x01 = vsli_n_u8(x0, x1, 3);
    uint16x8_t x01L = vmovl_u8(x01);

    uint32x4_t x01LL = vsraq_n_u32(vreinterpretq_u32_u16(x01L), vreinterpretq_u32_u16(x01L), 10);
    x01LL = vmovl_u16(vmovn_u32(x01LL));

    uint64x2_t x01X = vsraq_n_u64(vreinterpretq_u64_u32(x01LL), vreinterpretq_u64_u32(x01LL), 20);
    x01X = vmovl_u32(vmovn_u64(x01X));

    uint64x1_t X0 = vget_low_u64(x01X);
    uint64x1_t X1 = vget_high_u64(x01X);
    X0 = vsli_n_u64(X0, X1, 24);
    vst1_u32(dst, vreinterpret_u32_u64(X0));
}