通过创建单个线程来运行一段代码,从而加快执行速度

时间:2019-06-23 13:20:02

标签: c++ multithreading pthreads simd avx2

我正在尝试使用Intel AVX2和Posix线程进行一些基准测试。 假设我正在尝试查找样本中的最小值。 当我创建一个简单的程序时,我将运行avx_min函数。 当我创建一个内部创建Posix线程的程序时,已将avx_min的实现更改为avx_min_thread,如下所示,但实际实现保持不变。此函数可用于多个线程,并且不需要同步,因为线程不会“冲突”(tid = 0、1、2等)。

当我编译两个实现而没有指定任何优化标志时,它们给我两个相同的时间结果。另一个大小,当我使用-O3标志进行编译时,它们导致不同的执行时间,而我无法弄清楚为什么会发生这种情况。

P.S:我使用以下命令对其进行编译:

  • 情况1(不创建线程):g ++ -mavx2 -O3 -o avxMinO3 avxMinO3.cpp

  • 案例2(在内部创建posix线程):g ++ -mavx2 avxMinO3_t.cpp -lpthread -O3 -o avxMinO3_t

P.S 2:

  • 第一例执行时间:0.34秒
  • 第二例执行时间:0.049秒
Case 1:

double initialize_input(int32_t** relation, int32_t value_bound, int32_t input_size){

  clock_t t;
  srand(time(NULL));
  t = clock();
  for(int32_t i = 0 ; i < input_size ; i++){
    (*relation)[i] = rand() % value_bound;
  }
  t = clock() - t;
  return ((double) t) / CLOCKS_PER_SEC;
}


int* avx_min(int32_t** relation, int32_t rel_size, double* function_time){

  clock_t tic, tac;
  __m256i input_buffer;
  int32_t* rel = (*relation);
  __m256i min = _mm256_set_epi32(INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX);
  tic = clock();
  for(int i = 0 ; i < rel_size ; i += 8){
    input_buffer = _mm256_stream_load_si256((__m256i*)&rel[i]);
    min = _mm256_min_epi32(min, input_buffer);
  }
  tac = clock();
  double time_diff = (double)(tac - tic);
  (*function_time) = time_diff / CLOCKS_PER_SEC;
  int* temp = (int*)&min;
  return temp;
}


int main(int argc, char** argv) {

  int32_t* relation;
  double* function_time;
  int32_t input_size = 1024 * 1024 * 1024;
  int32_t value_bound = 1000;

  int alloc_time = initialize_input(&relation, value_bound, input_size);
  int* res = avx_min(&relation, input_size, function_time);
  return 0;
}

情况2:

template<typename T>
struct thread_input {
  T* relation;
  T rel_size;
  double function_time;
  short numberOfThreads;
  short tid;
};


template<typename T, typename S, typename I, typename RELTYPE>
T** createAndInitInputPtr(S numberOfThreads, I rel_size, S value_bound, RELTYPE** relation ){
  T **result = new T*[numberOfThreads];

  for (int i = 0; i < numberOfThreads; i++) {
    result[i] = new T;
    result[i]->rel_size = rel_size;
    result[i]->relation = (*relation);
    result[i]->numberOfThreads = numberOfThreads;
    result[i]->tid = i;
  }
  return result;
}


void* avx_min_t(void* input){

    clock_t tic, tac;
    struct thread_input<int32_t> *input_ptr;
    input_ptr = (struct thread_input<int32_t>*) input;

    int32_t* relation = input_ptr->relation;
    int32_t rel_size = input_ptr->rel_size;
    int32_t start = input_ptr->tid * 8;
    int32_t offset = input_ptr->numberOfThreads * 8;

    __m256i input_buffer;
    __m256i min = _mm256_set_epi32(INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX, INT32_MAX);
    tic = clock();
    for(int i = start ; i < rel_size ; i += offset){
      input_buffer = _mm256_stream_load_si256((__m256i*)&relation[i]);
      min = _mm256_min_epi32(min, input_buffer);
    }
    tac = clock();
    double time_diff = (double)(tac-tic);
    time_diff = time_diff / CLOCKS_PER_SEC;
    input_ptr->function_time = time_diff;
}


int main(int argc, char* argv[]){

  int rel_size = 1024 * 1024 * 1024;
  short numberOfThreads = 1;
  short value_bound = 1000;

  pthread_t* threads = new pthread_t[numberOfThreads];

  short flag = 1; // flag to check proper aligned memory allocations

  int32_t* relation;
  double alloc_time = 0.0;
  flag = posix_memalign((void**)&relation, 32, rel_size * sizeof(int32_t));
  if(flag) {
    std::cout << "Memory allocation problem. Exiting..." << std::endl;
    exit(1);
  }
  alloc_time += initialize_input(&relation, value_bound, rel_size);


  struct thread_input<int32_t> **input_ptr = createAndInitInputPtr<struct thread_input<int32_t>, short, int, int32_t>(numberOfThreads, rel_size, value_bound, &relation);

  clock_t tic = clock();

  for (int i = 0; i < numberOfThreads; i++) {
    pthread_create(&threads[i], NULL, avx_min_t,(void*) input_ptr[i]);
  }


  for (int i = 0; i < numberOfThreads; i++) {
    pthread_join(threads[i], NULL);
  }
  tic = clock()-tic;
  double time = tic / CLOCKS_PER_SEC;
  std::cout << time << std::endl;
  return 0;
}

1 个答案:

答案 0 :(得分:3)

void* avx_min_t(void* input)min无效,因此可以优化从阵列加载的SIMD工作。

其内部循环使用gcc -O3 -march=haswell进行编译,并且clang基本相同。

.L3:
        add     ebx, r12d
        cmp     r13d, ebx
        jg      .L3

因此,它实际上只是asm中的一个空循环,需要0.04秒的时间将指针增加4GB / 32 bytes次。

for(int i = start ; i < rel_size ; i += offset){
}

我认为您打算返回某项内容,因为该函数已声明为void*,并且具有从非void函数的结尾掉下来的不确定行为。 默认情况下,GCC和clang会发出警告,甚至不需要-Wall https://godbolt.org/z/Z1GWpU

<source>: In function 'void* avx_min_t(void*)':
<source>:66:1: warning: no return statement in function returning non-void [-Wreturn-type]
   66 | }
      | ^

始终检查编译器警告,尤其是在代码行为异常时。启用-Wall并修复所有警告。