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我正在尝试使用霓虹内在函数将交流代码转换为优化的代码。

这是对 2 个操作符而不是对操作符向量进行操作的 c 代码。

uint16_t mult_z216(uint16_t a,uint16_t b){
unsigned int c1 = a*b;
    if(c1)
    {
        int c1h = c1 >> 16;
        int c1l = c1 & 0xffff;
        return (c1l - c1h + ((c1l<c1h)?1:0)) & 0xffff;
    }
    return (1-a-b) & 0xffff;
}

此操作的 SEE 优化版本已通过以下方式实现：

#define MULT_Z216_SSE(a, b, c) \
    t0  = _mm_or_si128 ((a), (b)); \ //Computes the bitwise OR of the 128-bit value in a and the 128-bit value in b.
    (c) = _mm_mullo_epi16 ((a), (b)); \ //low 16-bits of the product of two 16-bit integers
    (a) = _mm_mulhi_epu16 ((a), (b)); \ //high 16-bits of the product of two 16-bit unsigned integers
    (b) = _mm_subs_epu16((c), (a)); \ //Subtracts the 8 unsigned 16-bit integers of a from the 8 unsigned 16-bit integers of c and saturates
    (b) = _mm_cmpeq_epi16 ((b), C_0x0_XMM); \ //Compares the 8 signed or unsigned 16-bit integers in a and the 8 signed or unsigned 16-bit integers in b for equality. (0xFFFF or 0x0)
    (b) = _mm_srli_epi16 ((b), 15); \ //shift right 16 bits
    (c) = _mm_sub_epi16 ((c), (a)); \ //Subtracts the 8 signed or unsigned 16-bit integers of b from the 8 signed or unsigned 16-bit integers of a.
    (a) = _mm_cmpeq_epi16 ((c), C_0x0_XMM); \ ////Compares the 8 signed or unsigned 16-bit integers in a and the 8 signed or unsigned 16-bit integers in b for equality. (0xFFFF or 0x0)
    (c) = _mm_add_epi16 ((c), (b)); \ // Adds the 8 signed or unsigned 16-bit integers in a to the 8 signed or unsigned 16-bit integers in b.
    t0  = _mm_and_si128 (t0, (a)); \ //Computes the bitwise AND of the 128-bit value in a and the 128-bit value in b.
    (c) = _mm_sub_epi16 ((c), t0); ///Subtracts the 8 signed or unsigned 16-bit integers of b from the 8 signed or unsigned 16-bit integers of a.

我几乎已经使用霓虹内在函数转换了这个：

#define MULT_Z216_NEON(a, b, out) \
    temp = vorrq_u16 (*a, *b); \
    // ??
    // ??
    *b = vsubq_u16(*out, *a); \
    *b = vceqq_u16(*out, vdupq_n_u16(0x0000)); \
    *b = vshrq_n_u16(*b, 15); \
    *out = vsubq_s16(*out, *a); \
    *a = vceqq_s16(*c, vdupq_n_u16(0x0000)); \
    *c = vaddq_s16(*c, *b); \
    *temp = vandq_u16(*temp, *a); \
    *out = vsubq_s16(*out, *a);

我只是错过了和的霓虹灯等价_mm_mullo_epi16 ((a), (b));物_mm_mulhi_epu16 ((a), (b));。要么我误解了某些东西，要么 NEON 中没有这样的内在函数。如果没有等效的如何使用 NEONS 内在函数归档这些步骤？

更新 ：

我忘了强调以下一点：函数的操作符是uint16x8_t NEON 向量（每个元素都是 uint16_t => 0 到 65535 之间的整数）。在一个答案中，有人建议使用内在的vqdmulhq_s16(). 使用这个与给定的实现不匹配，因为乘法内在函数会将向量解释为有符号值并产生错误的输出。

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2 回答 2

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您可以使用：

uint32x4_t vmull_u16 (uint16x4_t, uint16x4_t)

它返回一个 32 位乘积的向量。如果您想将结果分成高低部分，您可以使用 NEON unzip 内在函数。

于 2012-07-02T18:30:50.593 回答
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vmulq_s16() 相当于 _mm_mullo_epi16。_mm_mulhi_epu16 没有完全等价的；最接近的指令是 vqdmulhq_s16() ，它是“饱和、加倍、乘法、返回高位部分”。它仅对有符号的 16 位值进行操作，您需要将输入或输出除以 2 以消除加倍。

于 2012-07-02T22:02:13.933 回答