Question

_mm_avg_epu16通过PAVGW提供平均两个无符号16位整数。转换为float并除以2.唯一合适的（最佳）方式使用SSE来获得~~平均两个带符号的16位整数~~一个“有符号的平均轮次然后反转顶部位“（@Mysticial），还是有另一种方式？

编辑：这是我想要优化的代码，到目前为止我使用SSE的所有尝试都已接近但不完全匹配，通常是围绕饱和/溢出包装的问题：

int16_t *a;
int16_t *b;
uint16_t *out;

out[i] = int((a[i] + b[i]) / 2.0f + 32768.5f)

尝试＃1：

const __m128i outputVal = _mm_add_epi16(_mm_avg_epu16(a, b),  _mm_set1_epi16(32768));

尝试＃2：

const __m128i sum = _mm_add_epi16(a, b);
const __m128i outputVal = _mm_add_epi16(_mm_srai_epi16(sum, 1), _mm_set1_epi16(32768));

尝试＃3：

const __m128 elt_offset = _mm_set1_ps(32768.5f);

const __m128 avg_divisor = _mm_set1_ps(2.f);

const __m128i eltSum = _mm_add_epi16(edgeRowElts, edgeInnerRowElts); /* eltSum = int((inputData[i] + inputData[i + (direction*x)]) */
const __m64 eltSumLow  = _mm_movepi64_pi64(eltSum); /* eltSumLow = (__m64) (0x0ffffffff & eltSum) */
const __m64 eltSumHigh = _mm_movepi64_pi64(_mm_srli_si128(eltSum, 8)); /* eltSumHigh = (__m64) (0x0ffffffff & (eltSum >> 64)) */

/* Lower */
__m128 eltSumF = _mm_cvtpi16_ps(eltSumLow); /* eltSumF = (float) eltSum; */

__m128 eltAvg  = _mm_div_ps(eltSumF, avg_divisor); /* eltAvg = eltSum / 2.0f */
__m128 eltAvgOffset = _mm_add_ps(eltAvg,  elt_offset); /* eltAvgOffset = eltAvg + 32768.5f */
const __m64 outputValLow  = _mm_cvtps_pi16(eltAvgOffset); /* outputVal = (short) eltAvgOffset */

/* Upper */
eltSumF = _mm_cvtpi16_ps(eltSumHigh); /* eltSumF = (float) eltSum; */

eltAvg  = _mm_div_ps(eltSumF, avg_divisor); /* eltAvg = eltSum / 2.0f */
eltAvgOffset = _mm_add_ps(eltAvg,  elt_offset); /* eltAvgOffset = eltAvg + 32768.5f */
const __m64 outputValHigh = _mm_cvtps_pi16(eltAvgOffset); /* outputVal = (short) eltAvgOffset */

__m128i outputVal = _mm_slli_si128(_mm_movpi64_epi64(outputValHigh), 8); /* outputVal = (outputValHigh << 64); */
outputVal = _mm_or_si128(outputVal, _mm_movpi64_epi64(outputValLow)); /* outputVal = outputVal | (outputValLow); */

Answer 1

我不确定我是否完全理解这里的所有要求，但似乎：

a = _mm_add_epi16(a, _mm_set1_epi16(32768));
b = _mm_add_epi16(b, _mm_set1_epi16(32768));
outputVal = _mm_avg_epu16(a, b);

除了整理要求外，

应该给你一切。

如果是这样的话，那么事后不应该很难修正四舍五入：

round = _mm_xor_si128(a, b);
round = _mm_and_si128(round, _mm_set1_epi16(1));
outputVal = _mm_add_epi16(outputVal, round);

签署了16位SSE平均值

1 个答案: