如何将含可变比特数的NEON向量归约为单个32位值？

需求说明 #

输入为两个NEON uint8x16_t SIMD寄存器：

• 第一个寄存器的每个元素低N位为有效信息，剩余8-N位为0；
• 第二个寄存器对应元素存储该N值，且所有元素的N之和≤32。

需将这些有效比特无填充地拼接为单个32位寄存器，可调整有效比特的存储位置（如改存高N位）或元素顺序，唯一硬限制是拼接无填充且顺序固定。

现有实现方案 #

1. 循环累加的朴素方案 #

uint32_t bit_reduce_naive(uint8x16_t bits, uint8x16_t bit_counts)
{
    uint32_t result = 0;
    for (size_t i = 0; i != 16; ++i)
    {
        result <<= bit_counts[i];
        result |= bits[i];
    }
    return result;
}

2. 迭代合并的SIMD优化方案 #

先实现u8转u16的归约步骤：

inline std::pair<uint16x8_t, uint16x8_t> reduce_step_u8(uint8x16_t bits, uint8x16_t bit_counts)
{
    const uint16x8_t expanded_count_lower = vmovl_u8(vget_low_u8(bit_counts));
    const uint16x8_t expanded_bits_lower  = vmovl_u8(vget_low_u8(bits));
    const uint16x8_t expanded_bits_higher = vmovl_high_u8(bits);
    return std::pair
    {
        vorrq_u16(expanded_bits_lower, vshlq_u16(expanded_bits_higher, expanded_count_lower)),
        vaddl_u8( vget_low_u8(bit_counts), vget_high_u8(bit_counts)),
    };
}

再通过链式调用将输入归约为两个uint64_t值后合并：

uint32_t bit_reduce_better(uint8x16_t bits, uint8x16_t bit_counts)
{
    const auto [bits_u16, counts_u16] = reduce_step_u8(bits, bit_counts);
    const auto [bits_u32, counts_u32] = reduce_step_u16(bits_u16, counts_u16);
    const auto [bits_u64, counts_u64] = reduce_step_u32(bits_u32, counts_u32);

    const uint64_t low_bits  = vget_low_u64(bits_u64)[0];
    const uint64_t high_bits = vget_low_u64(bits_u64)[0]; // 注：原代码此处疑似笔误，应为vget_high_u64
    const uint64_t low_bits_count = vget_high_u64(counts_u64)[0];

    return static_cast<uint32_t>(low_bits | (high_bits << low_bits_count));
}

注：该SIMD方案的比特拼接顺序与原向量不同，但顺序并非硬要求。现有方案已满足功能需求，现寻求更高效的实现方案。

内容的提问来源于stack exchange，提问作者user19232978