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c++ - 这个 SSE2 转置有什么问题？

问题描述

我正在尝试转换此代码：

double *pB = b[voiceIndex];
double *pC = c[voiceIndex];
double phase = mPhase;
double bp0 = mNoteFrequency * mHostPitch;

for (int sampleIndex = 0; sampleIndex < blockSize; sampleIndex++) {
    // some other code (that will use phase, like sin(phase))

    phase += std::clamp(radiansPerSample * (bp0 * pB[sampleIndex] + pC[sampleIndex]), 0.0, PI);
}

mPhase = phase;

在 SSE2 中，试图加速整个区块（经常被调用）。我正在使用带有快速优化标志的 MSVC，但自动矢量化非常糟糕。因为我也在学习矢量化，所以我觉得这是一个很好的挑战。

所以我采用了上面的公式，并进行了简化，例如：

radiansPerSampleBp0 = radiansPerSample * bp0;
phase += std::clamp(radiansPerSampleBp0 * pB[sampleIndex] + radiansPerSample * pC[sampleIndex]), 0.0, PI);

可以将其静音为串行依赖项，例如：

phase[0] += (radiansPerSampleBp0 * pB[0] + radiansPerSample * pC[0])
phase[1] += (radiansPerSampleBp0 * pB[1] + radiansPerSample * pC[1]) + (radiansPerSampleBp0 * pB[0] + radiansPerSample * pC[0])

phase[2] += (radiansPerSampleBp0 * pB[2] + radiansPerSample * pC[2]) + (radiansPerSampleBp0 * pB[1] + radiansPerSample * pC[1])
phase[3] += (radiansPerSampleBp0 * pB[3] + radiansPerSample * pC[3]) + (radiansPerSampleBp0 * pB[2] + radiansPerSample * pC[2])

phase[4] += (radiansPerSampleBp0 * pB[4] + radiansPerSample * pC[4]) + (radiansPerSampleBp0 * pB[3] + radiansPerSample * pC[3])
phase[5] += (radiansPerSampleBp0 * pB[5] + radiansPerSample * pC[5]) + (radiansPerSampleBp0 * pB[4] + radiansPerSample * pC[4])

因此，我做的代码：

double *pB = b[voiceIndex];
double *pC = c[voiceIndex];
double phase = mPhase;
double bp0 = mNoteFrequency * mHostPitch;

__m128d v_boundLower = _mm_set1_pd(0.0);
__m128d v_boundUpper = _mm_set1_pd(PI);
__m128d v_radiansPerSampleBp0 = _mm_set1_pd(mRadiansPerSample * bp0);
__m128d v_radiansPerSample = _mm_set1_pd(mRadiansPerSample);

__m128d v_pB0 = _mm_load_pd(pB);
v_pB0 = _mm_mul_pd(v_pB0, v_radiansPerSampleBp0);
__m128d v_pC0 = _mm_load_pd(pC);
v_pC0 = _mm_mul_pd(v_pC0, v_radiansPerSample);

__m128d v_pB1 = _mm_setr_pd(0.0, pB[0]);
v_pB1 = _mm_mul_pd(v_pB1, v_radiansPerSampleBp0);
__m128d v_pC1 = _mm_setr_pd(0.0, pC[0]);
v_pC1 = _mm_mul_pd(v_pC1, v_radiansPerSample);

__m128d v_phase = _mm_set1_pd(phase);
__m128d v_phaseAcc;

for (int sampleIndex = 0; sampleIndex < blockSize; sampleIndex += 2, pB += 2, pC += 2) {
    // some other code (that will use phase, like sin(phase))

    v_phaseAcc = _mm_add_pd(v_pB0, v_pC0);
    v_phaseAcc = _mm_max_pd(v_phaseAcc, v_boundLower);
    v_phaseAcc = _mm_min_pd(v_phaseAcc, v_boundUpper);
    v_phaseAcc = _mm_add_pd(v_phaseAcc, v_pB1);
    v_phaseAcc = _mm_add_pd(v_phaseAcc, v_pC1);
    v_phase = _mm_add_pd(v_phase, v_phaseAcc);

    v_pB0 = _mm_load_pd(pB + 2);
    v_pB0 = _mm_mul_pd(v_pB0, v_radiansPerSampleBp0);
    v_pC0 = _mm_load_pd(pC + 2);
    v_pC0 = _mm_mul_pd(v_pC0, v_radiansPerSample);

    v_pB1 = _mm_load_pd(pB + 1);
    v_pB1 = _mm_mul_pd(v_pB1, v_radiansPerSampleBp0);
    v_pC1 = _mm_load_pd(pC + 1);
    v_pC1 = _mm_mul_pd(v_pC1, v_radiansPerSample);
}

mPhase = v_phase.m128d_f64[blockSize % 2 == 0 ? 1 : 0];

但是，不幸的是，在求和“步骤”之后，每个相位值的结果变得非常不同。尝试调试，但我无法真正找到问题所在。

此外，它并不是真的那么“快”而不是旧版本。

你能认清问题吗？以及如何加速代码？

这是整个代码，如果你想检查两个不同的输出：

#include <iostream>
#include <algorithm>
#include <immintrin.h>
#include <emmintrin.h>

#define PI 3.14159265358979323846

constexpr int voiceSize = 1;
constexpr int bufferSize = 256;

class Param
{
public:
    alignas(16) double mPhase = 0.0;
    alignas(16) double mPhaseOptimized = 0.0;
    alignas(16) double mNoteFrequency = 10.0;
    alignas(16) double mHostPitch = 1.0;
    alignas(16) double mRadiansPerSample = 1.0;

    alignas(16) double b[voiceSize][bufferSize];
    alignas(16) double c[voiceSize][bufferSize];

    Param() { }

    inline void Process(int voiceIndex, int blockSize) {
        double *pB = b[voiceIndex];
        double *pC = c[voiceIndex];
        double phase = mPhase;
        double bp0 = mNoteFrequency * mHostPitch;

        for (int sampleIndex = 0; sampleIndex < blockSize; sampleIndex++) {
            // some other code (that will use phase, like sin(phase))

            phase += std::clamp(mRadiansPerSample * (bp0 * pB[sampleIndex] + pC[sampleIndex]), 0.0, PI);

            std::cout << sampleIndex << ": " << phase << std::endl;
        }

        mPhase = phase;
    }
    inline void ProcessOptimized(int voiceIndex, int blockSize) {
        double *pB = b[voiceIndex];
        double *pC = c[voiceIndex];
        double phase = mPhaseOptimized;
        double bp0 = mNoteFrequency * mHostPitch;

        __m128d v_boundLower = _mm_set1_pd(0.0);
        __m128d v_boundUpper = _mm_set1_pd(PI);
        __m128d v_radiansPerSampleBp0 = _mm_set1_pd(mRadiansPerSample * bp0);
        __m128d v_radiansPerSample = _mm_set1_pd(mRadiansPerSample);

        __m128d v_pB0 = _mm_load_pd(pB);
        v_pB0 = _mm_mul_pd(v_pB0, v_radiansPerSampleBp0);
        __m128d v_pC0 = _mm_load_pd(pC);
        v_pC0 = _mm_mul_pd(v_pC0, v_radiansPerSample);

        __m128d v_pB1 = _mm_setr_pd(0.0, pB[0]);
        v_pB1 = _mm_mul_pd(v_pB1, v_radiansPerSampleBp0);
        __m128d v_pC1 = _mm_setr_pd(0.0, pC[0]);
        v_pC1 = _mm_mul_pd(v_pC1, v_radiansPerSample);

        __m128d v_phase = _mm_set1_pd(phase);
        __m128d v_phaseAcc;

        for (int sampleIndex = 0; sampleIndex < blockSize; sampleIndex += 2, pB += 2, pC += 2) {
            // some other code (that will use phase, like sin(phase))

            v_phaseAcc = _mm_add_pd(v_pB0, v_pC0);
            v_phaseAcc = _mm_max_pd(v_phaseAcc, v_boundLower);
            v_phaseAcc = _mm_min_pd(v_phaseAcc, v_boundUpper);
            v_phaseAcc = _mm_add_pd(v_phaseAcc, v_pB1);
            v_phaseAcc = _mm_add_pd(v_phaseAcc, v_pC1);
            v_phase = _mm_add_pd(v_phase, v_phaseAcc);

            v_pB0 = _mm_load_pd(pB + 2);
            v_pB0 = _mm_mul_pd(v_pB0, v_radiansPerSampleBp0);
            v_pC0 = _mm_load_pd(pC + 2);
            v_pC0 = _mm_mul_pd(v_pC0, v_radiansPerSample);

            v_pB1 = _mm_load_pd(pB + 1);
            v_pB1 = _mm_mul_pd(v_pB1, v_radiansPerSampleBp0);
            v_pC1 = _mm_load_pd(pC + 1);
            v_pC1 = _mm_mul_pd(v_pC1, v_radiansPerSample);

            std::cout << sampleIndex << ": " << v_phase.m128d_f64[0] << std::endl;
            std::cout << sampleIndex + 1 << ": " << v_phase.m128d_f64[1] << std::endl;
        }

        mPhaseOptimized = v_phase.m128d_f64[blockSize % 2 == 0 ? 1 : 0];
    }
};

class MyPlugin
{
public: 
    Param mParam1;

    MyPlugin() {
        // fill b
        for (int voiceIndex = 0; voiceIndex < voiceSize; voiceIndex++) {
            for (int sampleIndex = 0; sampleIndex < bufferSize; sampleIndex++) {
                double value = (sampleIndex / ((double)bufferSize - 1));

                mParam1.b[voiceIndex][sampleIndex] = value;
            }
        }

        // fill c
        for (int voiceIndex = 0; voiceIndex < voiceSize; voiceIndex++) {
            for (int sampleIndex = 0; sampleIndex < bufferSize; sampleIndex++) {
                double value = 0.0;

                mParam1.c[voiceIndex][sampleIndex] = value;
            }
        }
    }
    ~MyPlugin() { }

    void Process(int blockSize) {
        for (int voiceIndex = 0; voiceIndex < voiceSize; voiceIndex++) {
            mParam1.Process(voiceIndex, blockSize);
        }
    }
    void ProcessOptimized(int blockSize) {
        for (int voiceIndex = 0; voiceIndex < voiceSize; voiceIndex++) {
            mParam1.ProcessOptimized(voiceIndex, blockSize);
        }
    }
};

int main() {
    MyPlugin myPlugin;

    long long numProcessing = 1;
    long long counterProcessing = 0;

    // I'll only process once block, just for analysis
    while (counterProcessing++ < numProcessing) {
        // variable blockSize (i.e. it can vary, being even or odd)
        int blockSize = 256;

        // process data
        myPlugin.Process(blockSize);
        std::cout << "#########" << std::endl;
        myPlugin.ProcessOptimized(blockSize);
    }
}

标签： c++vectorizationintrinsicssse2

解决方案

（更新：这个答案是在显示在循环内使用的编辑之前写的v_phase。）

等一下，我认为在你之前的问题中你需要phase每一步的值。是 // some other code (that will use phase)的，循环中有一条评论。

但这看起来你只对最终值感兴趣。因此，您可以自由地重新排序，因为每个步骤的钳位都是独立的。

这只是一个缩减（如数组的总和），通过一些动态处理来生成缩减的输入。

您希望 2 个元素是v_phase偶数/奇数元素的 2 个独立部分和。然后你在最后水平求和。（例如_mm_unpackhi_pd(v_phase, v_phase)，将高半部分置于底部，或参见Fastest way to do Horizo​​ntal float vector sum on x86）。

然后可以选择fmod在结果上使用标量以将范围缩小到[0..2Pi)范围内。（在求和过程中偶尔缩小范围可以通过阻止值变得如此之大来帮助精度，如果结果证明精度成为问题的话。）

如果不是这种情况，并且您{ phase[i+0], phase[i+1] }在每i+=2一步都需要一个 for something 的向量，那么您的问题似乎与前缀 sum相关。但是每个向量只有 2 个元素，只是冗余地对具有未对齐负载的元素执行所有操作可能是有意义的。

节省的成本可能比我想象的要少，因为您需要分别钳制每个步骤：pB[i+0] + pB[i+1]在乘法之前执行可能会导致不同的钳制。

但是您显然已经删除了我们简化公式中的限制，因此您可以在应用 mul/add 公式之前添加元素。

或者，一次执行两个步骤的乘法/加法可能是一种胜利，然后随机播放以添加正确的内容。

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