0

我一直在使用 altivec 来实现基本的数学运算，作为为即将到来的项目学习 simd 的一种方式。此外，作为查看它的性能优势的一种方式，我跟踪执行操作需要多长时间，但我遇到了一些奇怪的事情。

我做的第一件事是将两个向量相加并减去两个向量。这工作正常。我做的下一件事是将两个向量相乘。然而，乘法比加法更快，即使根据我的特定 CPU 数据表中关于所使用指令的内容，使用更少的时钟周期来添加乘法。

我有两个每个 10MB 大的数组，并通过这两个例程运行它们：

void av_AddValues(int32_t* intArrayA, int32_t* intArrayB, int32_t* outputBuffer, int size)
{
  int iterations = size / (sizeof(__vector int32_t) / sizeof(int32_t));

  __vector int32_t* tempA = (__vector int32_t *) intArrayA;
  __vector int32_t* tempB = (__vector int32_t *) intArrayB;
  __vector int32_t* tempOut = (__vector int32_t *) outputBuffer;
  for(int i = 0; i < iterations; i++)
  {
    __vector int32_t sum = vec_add(*tempA, *tempB);
    vec_st(sum, 0, tempOut);

    tempA++;
    tempB++;
    tempOut++;
  }
}

  void av_MultiplyValues(int16_t* intArrayA, int16_t* intArrayB, int32_t* outputBuffer, int size)
  {
    int iterations = size / (sizeof(__vector int16_t) / sizeof(int16_t));
    __vector int16_t* tempA = (__vector int16_t *) intArrayA;
    __vector int16_t* tempB = (__vector int16_t *) intArrayB;
    __vector int32_t* tempOut = (__vector int32_t *) outputBuffer;


    for(int i = 0; i < iterations; i++)
    {
      __vector int32_t productEven = vec_mule(*tempA, *tempB);
      __vector int32_t productOdd = vec_mulo(*tempA, *tempB);

      __vector int32_t mergedProductHigh = vec_mergeh(productEven, productOdd);
      __vector int32_t mergedProductLow = vec_mergel(productEven, productOdd);

      vec_st(mergedProductHigh, 0, tempOut);
      tempOut++;
      vec_st(mergedProductLow, 0, tempOut);

      tempA++;
      tempB++;
      tempOut++;
    }
  }

在我的特定平台上，处理 av_AddValues 需要 81 毫秒，处理 av_MultiplyValues 需要 48 毫秒。（使用 std::chrono::high_resolution_clock 记录的时间）

为什么乘法比加法花费更少的时间来处理？

考虑到 __vector 类型总是处理 16 字节的数据，我认为添加 32 位值与乘以 16 位值并没有什么不同。

我的第一个想法是，由于将数字加在一起是一项微不足道的任务，CPU 完成操作的速度比它从内存中获取数据的速度要快。而对于乘法，这种获取延迟被 CPU 忙于工作并且永远不必等待这么久的事实所掩盖。

这是一个正确的假设吗？

完整代码：

#include <chrono>
#include <random>
#include <limits>

#include <iostream>
#include <cassert>
#include <cstring>
#include <cstdint>
#include <malloc.h>

#include <altivec.h>
#undef vector

void GenerateRandom16bitValues(int16_t* inputABuffer, int16_t* inputBBuffer, int32_t* outputBuffer, int size);
void GenerateRandom32bitValues(int32_t* inputABuffer, int32_t* inputBBuffer, int32_t* outputBuffer, int size);
void TestAdd();
void TestMultiply();
void av_AddValues(int32_t* intArrayA, int32_t* intArrayB, int32_t* outputBuffer, int size);
void av_MultiplyValues(int16_t* intArrayA, int16_t* intArrayB, int32_t* outputBuffer, int size);

int main()
{
  TestAdd();
  TestMultiply();
}

void GenerateRandom16bitValues(int16_t* inputABuffer, int16_t* inputBBuffer, int32_t* outputBuffer, int size)
{
  std::random_device rd;
  std::mt19937 gen(rd());
  std::uniform_int_distribution<> dis(std::numeric_limits<int16_t>::min(), std::numeric_limits<int16_t>::max());

  for(int i = 0; i < size; i++)
  {
    inputABuffer[i] = dis(gen);
    inputBBuffer[i] = dis(gen);
    outputBuffer[i] = 0;
  }
}

void GenerateRandom32bitValues(int32_t* inputABuffer, int32_t* inputBBuffer, int32_t* outputBuffer, int size)
{
  std::random_device rd;
  std::mt19937 gen(rd());
  std::uniform_int_distribution<> dis(std::numeric_limits<int32_t>::min(), std::numeric_limits<int32_t>::max());

  for(int i = 0; i < size; i++)
  {
    inputABuffer[i] = dis(gen);
    inputBBuffer[i] = dis(gen);
    outputBuffer[i] = 0;
  }
}

void TestAdd()
{
    int size = 10'485'760;
    int bytes = size * sizeof(int32_t);

    int32_t* inputABuffer = (int32_t*) memalign(64, bytes);
    int32_t* inputBBuffer = (int32_t*) memalign(64, bytes);
    int32_t* outputBuffer = (int32_t*) memalign(64, bytes);
    assert(inputABuffer != nullptr);
    assert(inputBBuffer != nullptr);
    assert(outputBuffer != nullptr);

    GenerateRandom32bitValues(inputABuffer, inputBBuffer, outputBuffer, size);

    for(int i = 0; i < 20; i++)
    {
      auto start = std::chrono::high_resolution_clock::now();
      av_AddValues(inputABuffer, inputBBuffer, outputBuffer, size);
      auto end = std::chrono::high_resolution_clock::now();
      auto diff = std::chrono::duration_cast<std::chrono::milliseconds>(end - start);

      for(int k = 0; k < size; k++)
      {
        assert(outputBuffer[k] == (inputABuffer[k] + inputBBuffer[k]));
      }

      std::cout << "Vector Sum - " << diff.count() << "ms\n";
      memset(outputBuffer, 0, size);
    }
}

void TestMultiply()
{
    int size = 10'485'760;
    int16_t* inputABuffer = (int16_t*) memalign(64, size * sizeof(int16_t));
    int16_t* inputBBuffer = (int16_t*) memalign(64, size * sizeof(int16_t));
    int32_t* outputBuffer = (int32_t*) memalign(64, size * sizeof(int32_t));
    assert(inputABuffer != nullptr);
    assert(inputBBuffer != nullptr);
    assert(outputBuffer != nullptr);

    GenerateRandom16bitValues(inputABuffer, inputBBuffer, outputBuffer, size);

    for(int i = 0; i < 20; i++)
    {
      auto start = std::chrono::high_resolution_clock::now();
      av_MultiplyValues(inputABuffer, inputBBuffer, outputBuffer, size);
      auto end = std::chrono::high_resolution_clock::now();
      auto diff = std::chrono::duration_cast<std::chrono::milliseconds>(end - start);

      for(int k = 0; k < size; k++)
      {
        assert(outputBuffer[k] == (inputABuffer[k] * inputBBuffer[k]));
      }

      std::cout << "Vector product - " << diff.count() << "ms\n";
      memset(outputBuffer, 0, size);
    }
}

void av_AddValues(int32_t* intArrayA, int32_t* intArrayB, int32_t* outputBuffer, int size)
{
  int iterations = size / (sizeof(__vector int32_t) / sizeof(int32_t));

  __vector int32_t* tempA = (__vector int32_t *) intArrayA;
  __vector int32_t* tempB = (__vector int32_t *) intArrayB;
  __vector int32_t* tempOut = (__vector int32_t *) outputBuffer;

  for(int i = 0; i < iterations; i++)
  {
    __vector int32_t sum = vec_add(*tempA, *tempB);
    vec_st(sum, 0, tempOut);

    tempA++;
    tempB++;
    tempOut++;
  }
}

void av_MultiplyValues(int16_t* intArrayA, int16_t* intArrayB, int32_t* outputBuffer, int size)
{
  int iterations = size / (sizeof(__vector int16_t) / sizeof(int16_t));
  __vector int16_t* tempA = (__vector int16_t *) intArrayA;
  __vector int16_t* tempB = (__vector int16_t *) intArrayB;
  __vector int32_t* tempOut = (__vector int32_t *) outputBuffer;
  for(int i = 0; i < iterations; i++)
  {
    __vector int32_t productEven = vec_mule(*tempA, *tempB);
    __vector int32_t productOdd = vec_mulo(*tempA, *tempB);

    __vector int32_t mergedProductHigh = vec_mergeh(productEven, productOdd);
    __vector int32_t mergedProductLow = vec_mergel(productEven, productOdd);

    vec_st(mergedProductHigh, 0, tempOut);
    tempOut++;
    vec_st(mergedProductLow, 0, tempOut);

    tempA++;
    tempB++;
    tempOut++;
  }
}

perf stat 和 perf 记录的输出：

  Adding
   Performance counter stats for './alti':

         2151.146080      task-clock (msec)         #    0.999 CPUs utilized          
                   9      context-switches          #    0.004 K/sec                  
                   0      cpu-migrations            #    0.000 K/sec                  
               30957      page-faults               #    0.014 M/sec                  
          3871497132      cycles                    #    1.800 GHz                    
     <not supported>      stalled-cycles-frontend  
     <not supported>      stalled-cycles-backend   
          1504538891      instructions              #    0.39  insns per cycle        
           234038234      branches                  #  108.797 M/sec                  
              687912      branch-misses             #    0.29% of all branches        
           270305159      L1-dcache-loads           #  125.656 M/sec                  
            79819113      L1-dcache-load-misses     #   29.53% of all L1-dcache hits  
     <not supported>      LLC-loads                
     <not supported>      LLC-load-misses          

         2.152697186 seconds time elapsed


  CPU Utilization
    76.04%  alti     alti                 [.] av_AddValues    

  Multiply

  Performance counter stats for './alti':

         1583.016640      task-clock (msec)         #    0.999 CPUs utilized          
                   4      context-switches          #    0.003 K/sec                  
                   0      cpu-migrations            #    0.000 K/sec                  
               20717      page-faults               #    0.013 M/sec                  
          2849050875      cycles                    #    1.800 GHz                    
     <not supported>      stalled-cycles-frontend  
     <not supported>      stalled-cycles-backend   
          1520409634      instructions              #    0.53  insns per cycle        
           179185029      branches                  #  113.192 M/sec                  
              535437      branch-misses             #    0.30% of all branches        
           205341530      L1-dcache-loads           #  129.715 M/sec                  
            27124936      L1-dcache-load-misses     #   13.21% of all L1-dcache hits  
     <not supported>      LLC-loads                
     <not supported>      LLC-load-misses          

         1.584145737 seconds time elapsed


  CPU Utilization
    60.35%  alti     alti               [.] av_MultiplyValues
4

1 回答 1

2

它与输入缓冲区的大小有关。

在一种情况下（TestAdd）：

int size = 10'485'760;
int bytes = size * sizeof(int32_t);

int32_t* inputABuffer = (int32_t*) memalign(64, bytes);
int32_t* inputBBuffer = (int32_t*) memalign(64, bytes);
int32_t* outputBuffer = (int32_t*) memalign(64, bytes);

你分配 3 * size * 4 字节 (sizeof(int32_t) = 4)

在另一个 (test_mul) 中：

int size = 10'485'760;
int16_t* inputABuffer = (int16_t*) memalign(64, size * sizeof(int16_t));
int16_t* inputBBuffer = (int16_t*) memalign(64, size * sizeof(int16_t));
int32_t* outputBuffer = (int32_t*) memalign(64, size * sizeof(int32_t));

你分配 size*4 + 2*size*2 (sizeof(int16_t) = 2)

由于此代码完全受内存限制，因此第二个代码是 (3*4) / (4 + 2*2) = 1.5x faster。

这与您的测量结果一致，因为 2.15 / 1.5 = 1.43，接近 1.58。

于 2017-06-01T14:27:04.000 回答