C中最快的交错操作？

Question

我有一个指针的字节数组mixed,包含两个不同的阵列的交错字节array1和array2.说mixed看起来像这样:

a1b2c3d4...

我需要做的是解交织字节,所以我得到array1 = abcd...和array2 = 1234....我知道mixed提前的长度,长度array1和array2等价,都等于mixed / 2.

这是我目前的执行(array1和array2已经被分配):

int i, j;
int mixedLength_2 = mixedLength / 2;
for (i = 0, j = 0; i < mixedLength_2; i++, j += 2)
{
    array1[i] = mixed[j];
    array2[i] = mixed[j+1];
}

这避免了任何昂贵的乘法或除法运算,但仍然运行得不够快.我希望有类似的东西memcpy需要一个可以使用低级块复制操作来加速该过程的索引器.是否有比我现有的更快的实施？

编辑

目标平台是针对iOS和Mac的Objective-C.对于iOS设备而言,快速操作更为重要,因此针对iOS的解决方案将比没有更好.

更新

感谢大家的回应,尤其是Stephen Canon,Graham Lee和Mecki.这是我的"主"功能,使用Stephen的NEON内在函数(如果可用)和Graham的联合游标,Mecki建议的迭代次数减少.

void interleave(const uint8_t *srcA, const uint8_t *srcB, uint8_t *dstAB, size_t dstABLength)
{
#if defined __ARM_NEON__
    // attempt to use NEON intrinsics

    // iterate 32-bytes at a time
    div_t dstABLength_32 = div(dstABLength, 32);
    if (dstABLength_32.rem == 0)
    {
        while (dstABLength_32.quot --> 0)
        {
            const uint8x16_t a = vld1q_u8(srcA);
            const uint8x16_t b = vld1q_u8(srcB);
            const uint8x16x2_t ab = { a, b };
            vst2q_u8(dstAB, ab);
            srcA += 16;
            srcB += 16;
            dstAB += 32;
        }
        return;
    }

    // iterate 16-bytes at a time
    div_t dstABLength_16 = div(dstABLength, 16);
    if (dstABLength_16.rem == 0)
    {
        while (dstABLength_16.quot --> 0)
        {
            const uint8x8_t a = vld1_u8(srcA);
            const uint8x8_t b = vld1_u8(srcB);
            const uint8x8x2_t ab = { a, b };
            vst2_u8(dstAB, ab);
            srcA += 8;
            srcB += 8;
            dstAB += 16;
        }
        return;
    }
#endif

    // if the bytes were not aligned properly
    // or NEON is unavailable, fall back to
    // an optimized iteration

    // iterate 8-bytes at a time
    div_t dstABLength_8 = div(dstABLength, 8);
    if (dstABLength_8.rem == 0)
    {
        typedef union
        {
            uint64_t wide;
            struct { uint8_t a1; uint8_t b1; uint8_t a2; uint8_t b2; uint8_t a3; uint8_t b3; uint8_t a4; uint8_t b4; } narrow;
        } ab8x8_t;

        uint64_t *dstAB64 = (uint64_t *)dstAB;
        int j = 0;
        for (int i = 0; i < dstABLength_8.quot; i++)
        {
            ab8x8_t cursor;
            cursor.narrow.a1 = srcA[j  ];
            cursor.narrow.b1 = srcB[j++];
            cursor.narrow.a2 = srcA[j  ];
            cursor.narrow.b2 = srcB[j++];
            cursor.narrow.a3 = srcA[j  ];
            cursor.narrow.b3 = srcB[j++];
            cursor.narrow.a4 = srcA[j  ];
            cursor.narrow.b4 = srcB[j++];
            dstAB64[i] = cursor.wide;
        }
        return;
    }

    // iterate 4-bytes at a time
    div_t dstABLength_4 = div(dstABLength, 4);
    if (dstABLength_4.rem == 0)
    {
        typedef union
        {
            uint32_t wide;
            struct { uint8_t a1; uint8_t b1; uint8_t a2; uint8_t b2; } narrow;
        } ab8x4_t;

        uint32_t *dstAB32 = (uint32_t *)dstAB;
        int j = 0;
        for (int i = 0; i < dstABLength_4.quot; i++)
        {
            ab8x4_t cursor;
            cursor.narrow.a1 = srcA[j  ];
            cursor.narrow.b1 = srcB[j++];
            cursor.narrow.a2 = srcA[j  ];
            cursor.narrow.b2 = srcB[j++];
            dstAB32[i] = cursor.wide;
        }
        return;
    }

    // iterate 2-bytes at a time
    div_t dstABLength_2 = div(dstABLength, 2);
    typedef union
    {
        uint16_t wide;
        struct { uint8_t a; uint8_t b; } narrow;
    } ab8x2_t;

    uint16_t *dstAB16 = (uint16_t *)dstAB;
    for (int i = 0; i < dstABLength_2.quot; i++)
    {
        ab8x2_t cursor;
        cursor.narrow.a = srcA[i];
        cursor.narrow.b = srcB[i];
        dstAB16[i] = cursor.wide;
    }
}

void deinterleave(const uint8_t *srcAB, uint8_t *dstA, uint8_t *dstB, size_t srcABLength)
{
#if defined __ARM_NEON__
    // attempt to use NEON intrinsics

    // iterate 32-bytes at a time
    div_t srcABLength_32 = div(srcABLength, 32);
    if (srcABLength_32.rem == 0)
    {
        while (srcABLength_32.quot --> 0)
        {
            const uint8x16x2_t ab = vld2q_u8(srcAB);
            vst1q_u8(dstA, ab.val[0]);
            vst1q_u8(dstB, ab.val[1]);
            srcAB += 32;
            dstA += 16;
            dstB += 16;
        }
        return;
    }

    // iterate 16-bytes at a time
    div_t srcABLength_16 = div(srcABLength, 16);
    if (srcABLength_16.rem == 0)
    {
        while (srcABLength_16.quot --> 0)
        {
            const uint8x8x2_t ab = vld2_u8(srcAB);
            vst1_u8(dstA, ab.val[0]);
            vst1_u8(dstB, ab.val[1]);
            srcAB += 16;
            dstA += 8;
            dstB += 8;
        }
        return;
    }
#endif

    // if the bytes were not aligned properly
    // or NEON is unavailable, fall back to
    // an optimized iteration

    // iterate 8-bytes at a time
    div_t srcABLength_8 = div(srcABLength, 8);
    if (srcABLength_8.rem == 0)
    {
        typedef union
        {
            uint64_t wide;
            struct { uint8_t a1; uint8_t b1; uint8_t a2; uint8_t b2; uint8_t a3; uint8_t b3; uint8_t a4; uint8_t b4; } narrow;
        } ab8x8_t;

        uint64_t *srcAB64 = (uint64_t *)srcAB;
        int j = 0;
        for (int i = 0; i < srcABLength_8.quot; i++)
        {
            ab8x8_t cursor;
            cursor.wide = srcAB64[i];
            dstA[j  ] = cursor.narrow.a1;
            dstB[j++] = cursor.narrow.b1;
            dstA[j  ] = cursor.narrow.a2;
            dstB[j++] = cursor.narrow.b2;
            dstA[j  ] = cursor.narrow.a3;
            dstB[j++] = cursor.narrow.b3;
            dstA[j  ] = cursor.narrow.a4;
            dstB[j++] = cursor.narrow.b4;
        }
        return;
    }

    // iterate 4-bytes at a time
    div_t srcABLength_4 = div(srcABLength, 4);
    if (srcABLength_4.rem == 0)
    {
        typedef union
        {
            uint32_t wide;
            struct { uint8_t a1; uint8_t b1; uint8_t a2; uint8_t b2; } narrow;
        } ab8x4_t;

        uint32_t *srcAB32 = (uint32_t *)srcAB;
        int j = 0;
        for (int i = 0; i < srcABLength_4.quot; i++)
        {
            ab8x4_t cursor;
            cursor.wide = srcAB32[i];
            dstA[j  ] = cursor.narrow.a1;
            dstB[j++] = cursor.narrow.b1;
            dstA[j  ] = cursor.narrow.a2;
            dstB[j++] = cursor.narrow.b2;
        }
        return;
    }

    // iterate 2-bytes at a time
    div_t srcABLength_2 = div(srcABLength, 2);
    typedef union
    {
        uint16_t wide;
        struct { uint8_t a; uint8_t b; } narrow;
    } ab8x2_t;

    uint16_t *srcAB16 = (uint16_t *)srcAB;
    for (int i = 0; i < srcABLength_2.quot; i++)
    {
        ab8x2_t cursor;
        cursor.wide = srcAB16[i];
        dstA[i] = cursor.narrow.a;
        dstB[i] = cursor.narrow.b;
    }
}

Answer 1

在我的脑海中,我不知道用于解交织2个通道字节数据的库函数.但是,有必要向Apple提交错误报告以请求此类功能.

与此同时,使用NEON或SSE内在函数对这样的函数进行矢量化非常容易.具体来说,在ARM上,您将需要使用vld1q_u8从每个源数组加载向量,vuzpq_u8对它们进行去交织,以及vst1q_u8存储生成的向量; 这是一个粗略的草图,我没有测试过甚至试图建立,但它应该说明一般的想法.更复杂的实现肯定是可能的(特别是,NEON可以在一条指令中加载/存储两个 16B寄存器,编译器可能不会这样做,并且根据缓冲区的长度,一些流水线和/或展开可能是有益的是):

#if defined __ARM_NEON__
#   include <arm_neon.h>
#endif
#include <stdint.h>
#include <stddef.h>

void deinterleave(uint8_t *mixed, uint8_t *array1, uint8_t *array2, size_t mixedLength) {
#if defined __ARM_NEON__
    size_t vectors = mixedLength / 32;
    mixedLength %= 32;
    while (vectors --> 0) {
        const uint8x16_t src0 = vld1q_u8(mixed);
        const uint8x16_t src1 = vld1q_u8(mixed + 16);
        const uint8x16x2_t dst = vuzpq_u8(src0, src1);
        vst1q_u8(array1, dst.val[0]);
        vst1q_u8(array2, dst.val[1]);
        mixed += 32;
        array1 += 16;
        array2 += 16;
    }
#endif
    for (size_t i=0; i<mixedLength/2; ++i) {
        array1[i] = mixed[2*i];
        array2[i] = mixed[2*i + 1];
    }
}