等效于SSE本质的氖

Question

我正在尝试使用neon本质将c代码转换为优化代码。

这是c码，它的运算量超过两个运算子，而不是运算子的向量。

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;
}

以下操作已经实现了此操作的优化版本：

#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.

我几乎用霓虹灯的内在原理来改造这个：

#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);

我只想念_mm_mullo_epi16 ((a), (b));和_mm_mulhi_epu16 ((a), (b));的霓虹灯等价物。要么我误解了什么，要么霓虹灯里没有这样的本质。如果没有等效的，如何使用NEONS，本质，如何存档这些步骤？

最新情况：

我忘记强调以下一点:函数的运算子是，uint16x8_t，，NEON向量(每个元素是0到65535之间的uint16_t =>整数)。在一个答案中，有人建议使用固有的vqdmulhq_s16()。这种方法的使用与给定的实现不匹配，因为乘法内禀会将向量解释为有符号值，并产生错误的输出。

Answer 1

您可以使用：

uint32x4_t vmull_u16 (uint16x4_t, uint16x4_t) 

它返回32位产品的向量。如果你想把结果分解成高低部分，你可以使用霓虹灯解压缩内蕴。

Answer 2

vmulq_s16()相当于_mm_mullo_epi16。没有与_mm_mulhi_epu16完全等价的；最接近的指令是vqdmulhq_s16()，它是“饱和、倍、乘、返回高部分”。它只对有符号的16位值进行操作，您需要将输入或输出除以2才能使加倍无效。