3

我想使用流来并行化在单独的设备数据阵列上工作的内核的执行。数据在设备上分配并从以前的内核中填充。

我编写了以下程序，表明到目前为止我无法达到我的目标。实际上，两个非默认流上的内核在它们各自的流中顺序执行。

在 2 台装有最新 Debian linux 版本的 Intel 机器上观察到相同的行为。一个是带有 CUDA 4.2 的 Tesla C2075，另一个是带有 CUDA 5.0 的 Geforce 460GT。Visual Profiler 显示了 4.2 和 5.0 CUDA 版本中的顺序执行。

这是代码：

#include <iostream>
#include <stdio.h>
#include <ctime>

#include <curand.h>

using namespace std;

// compile and run this way:
// nvcc cuStreamsBasics.cu  -arch=sm_20  -o testCuStream   -lcuda -lcufft -lcurand 
// testCuStream  1024 512 512


/* -------------------------------------------------------------------------- */
//  "useful" macros
/* -------------------------------------------------------------------------- */


#define MSG_ASSERT( CONDITION, MSG )                    \
  if (! (CONDITION))                            \
    {                                   \
    std::cerr << std::endl << "Dynamic assertion `" #CONDITION "` failed in " << __FILE__ \
          << " line " << __LINE__ << ": <" << MSG << ">" << std::endl;  \
    exit( 1 );                              \
    } \ 



#define ASSERT( CONDITION ) \
  MSG_ASSERT( CONDITION, " " )



// allocate data on the GPU memory, unpinned
#define CUDALLOC_GPU( _TAB, _DIM, _DATATYPE ) \
  MSG_ASSERT( \
  cudaMalloc( (void**) &_TAB, _DIM * sizeof( _DATATYPE) ) \ 
== cudaSuccess , "failed CUDALLOC" );



/* -------------------------------------------------------------------------- */
// the CUDA kernels
/* -------------------------------------------------------------------------- */


// finds index in 1D array from sequential blocks
#define CUDAINDEX_1D                \
  blockIdx.y * ( gridDim.x * blockDim.x ) + \
  blockIdx.x * blockDim.x +         \
  threadIdx.x;                  \



__global__ void 
kernel_diva(float* data, float value, int array_size)
{
  int i = CUDAINDEX_1D
    if (i < array_size) 
      data[i] /= value;
}


__global__ void 
kernel_jokea(float* data, float value, int array_size)
{
  int i = CUDAINDEX_1D
    if (i < array_size) 
      data[i] *= value + sin( double(i)) * 1/ cos( double(i) );
}


/* -------------------------------------------------------------------------- */
// usage
/* -------------------------------------------------------------------------- */


static void
usage(int argc, char **argv) 
{
  if ((argc -1) != 3)
    {

      printf("Usage: %s <dimx> <dimy> <dimz> \n", argv[0]);
      printf("do stuff\n");

      exit(1);
    }  
}


/* -------------------------------------------------------------------------- */
// main program, finally!
/* -------------------------------------------------------------------------- */


int 
main(int argc, char** argv)
{
  usage(argc, argv);
  size_t x_dim = atoi( argv[1] );
  size_t y_dim = atoi( argv[2] );
  size_t z_dim = atoi( argv[3] );



  cudaStream_t stream1, stream2;
  ASSERT( cudaStreamCreate( &stream1 ) == cudaSuccess ); 
  ASSERT( cudaStreamCreate( &stream2 ) == cudaSuccess ); 



  size_t size = x_dim * y_dim * z_dim;
  float *data1, *data2;
  CUDALLOC_GPU( data1, size, float);
  CUDALLOC_GPU( data2, size, float);


  curandGenerator_t gen;
  curandCreateGenerator(&gen, CURAND_RNG_PSEUDO_DEFAULT);
  /* Set seed */
  curandSetPseudoRandomGeneratorSeed(gen, 1234ULL);
  /* Generate n floats on device */
  curandGenerateUniform(gen, data1, size);
  curandGenerateUniform(gen, data2, size);


  dim3 dimBlock( z_dim, 1, 1);
  dim3 dimGrid( x_dim, y_dim, 1);

  clock_t start;
  double diff;


  cudaDeviceSynchronize();
  start = clock(); 
  kernel_diva <<< dimGrid, dimBlock>>>( data1, 5.55f, size);   
  kernel_jokea<<< dimGrid, dimBlock>>>( data1, 5.55f, size);   
  kernel_diva <<< dimGrid, dimBlock>>>( data2, 5.55f, size); 
  kernel_jokea<<< dimGrid, dimBlock>>>( data2, 5.55f, size); 
  cudaDeviceSynchronize();
  diff = ( std::clock() - start ) / (double)CLOCKS_PER_SEC;

  cout << endl << "sequential: " << diff;


  cudaDeviceSynchronize();
  start = clock(); 
  kernel_diva <<< dimGrid, dimBlock, 0, stream1 >>>( data1, 5.55f, size); 
  kernel_diva <<< dimGrid, dimBlock, 0, stream2 >>>( data2, 5.55f, size); 
  kernel_jokea<<< dimGrid, dimBlock, 0, stream1 >>>( data1, 5.55f, size); 
  kernel_jokea<<< dimGrid, dimBlock, 0, stream2 >>>( data2, 5.55f, size); 
  cudaDeviceSynchronize();
  diff = ( std::clock() - start ) / (double)CLOCKS_PER_SEC;

  cout << endl << "parallel: " << diff;



  cudaStreamDestroy( stream1 ); 
  cudaStreamDestroy( stream2 ); 


  return 0;
}

通常，数组的维度是 s 512^3​​ingle float。我通常只是将数组切割成(512,1,1)我放在大小网格上的线程块(1<<15, (rest), 1)。

提前感谢您的任何提示或评论。

此致。

4

1 回答 1

5

我试图解释为什么您看不到两个内核的执行重叠。为此，我构建了下面报告的代码，它使用您的两个内核并监控每个块在哪个流式多处理器 (SM) 上运行。我正在使用 CUDA 6.5 (Release Candidate) 并且我在 GT540M 卡上运行，它只有2SM，因此它提供了一个简单的工作场所。blockSize选择权交给了新的 CUDA 6.5设施cudaOccupancyMaxPotentialBlockSize。

编码

#include <stdio.h>
#include <time.h>

//#define DEBUG_MODE

/********************/
/* CUDA ERROR CHECK */
/********************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
    if (code != cudaSuccess) 
    {
        fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
        if (abort) exit(code);
    }
}

/**************************************************/
/* STREAMING MULTIPROCESSOR IDENTIFICATION NUMBER */
/**************************************************/
__device__ unsigned int get_smid(void) {
    unsigned int ret;
    asm("mov.u32 %0, %smid;" : "=r"(ret) );
    return ret;
}

/************/
/* KERNEL 1 */
/************/
__global__ void kernel_1(float * __restrict__ data, const float value, int *sm, int N)
{
    int i = threadIdx.x + blockIdx.x * blockDim.x;

    if (i < N) {
        data[i] = data[i] / value;
        if (threadIdx.x==0) sm[blockIdx.x]=get_smid();
    }

}

//__global__ void kernel_1(float* data, float value, int N)
//{
//    int start = blockIdx.x * blockDim.x + threadIdx.x;
//    for (int i = start; i < N; i += blockDim.x * gridDim.x)
//    {
//        data[i] = data[i] / value; 
//    }
//}

/************/
/* KERNEL 2 */
/************/
__global__ void  kernel_2(float * __restrict__ data, const float value, int *sm, int N)
{
    int i = threadIdx.x + blockIdx.x*blockDim.x;

    if (i < N) {
        data[i] = data[i] * (value + sin(double(i)) * 1./cos(double(i)));
        if (threadIdx.x==0) sm[blockIdx.x]=get_smid();
    }
}

//__global__ void  kernel_2(float* data, float value, int N)
//{
//    int start = blockIdx.x * blockDim.x + threadIdx.x;
//    for (int i = start; i < N; i += blockDim.x * gridDim.x)
//    {
//        data[i] = data[i] * (value + sin(double(i)) * 1./cos(double(i)));
//    }
//}

/********/
/* MAIN */
/********/
int main()
{
    const int N = 10000;

    const float value = 5.55f;

    const int rep_num = 20;

    // --- CPU memory allocations
    float *h_data1 = (float*) malloc(N*sizeof(float));
    float *h_data2 = (float*) malloc(N*sizeof(float));
    float *h_data1_ref = (float*) malloc(N*sizeof(float));
    float *h_data2_ref = (float*) malloc(N*sizeof(float));

    // --- CPU data initializations
    srand(time(NULL));
    for (int i=0; i<N; i++) {
        h_data1[i] = rand() / RAND_MAX;
        h_data2[i] = rand() / RAND_MAX;
    }

    // --- GPU memory allocations
    float *d_data1, *d_data2;
    gpuErrchk(cudaMalloc((void**)&d_data1, N*sizeof(float)));
    gpuErrchk(cudaMalloc((void**)&d_data2, N*sizeof(float)));

    // --- CPU -> GPU memory transfers
    gpuErrchk(cudaMemcpy(d_data1, h_data1, N*sizeof(float), cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_data2, h_data2, N*sizeof(float), cudaMemcpyHostToDevice));

    // --- CPU data initializations
    srand(time(NULL));
    for (int i=0; i<N; i++) {
        h_data1_ref[i] = h_data1[i] / value;
        h_data2_ref[i] = h_data2[i] * (value + sin(double(i)) * 1./cos(double(i)));
    }

    // --- Stream creations
    cudaStream_t stream1, stream2;
    gpuErrchk(cudaStreamCreate(&stream1)); 
    gpuErrchk(cudaStreamCreate(&stream2)); 

    // --- Launch parameters configuration
    int blockSize1, blockSize2, minGridSize1, minGridSize2, gridSize1, gridSize2;
    cudaOccupancyMaxPotentialBlockSize(&minGridSize1, &blockSize1, kernel_1, 0, N); 
    cudaOccupancyMaxPotentialBlockSize(&minGridSize2, &blockSize2, kernel_2, 0, N); 

    gridSize1 = (N + blockSize1 - 1) / blockSize1; 
    gridSize2 = (N + blockSize2 - 1) / blockSize2; 

    // --- Allocating space for SM IDs
    int *h_sm_11 = (int*) malloc(gridSize1*sizeof(int)); 
    int *h_sm_12 = (int*) malloc(gridSize1*sizeof(int));
    int *h_sm_21 = (int*) malloc(gridSize2*sizeof(int));
    int *h_sm_22 = (int*) malloc(gridSize2*sizeof(int));
    int *d_sm_11, *d_sm_12, *d_sm_21, *d_sm_22;
    gpuErrchk(cudaMalloc((void**)&d_sm_11, gridSize1*sizeof(int)));
    gpuErrchk(cudaMalloc((void**)&d_sm_12, gridSize1*sizeof(int)));
    gpuErrchk(cudaMalloc((void**)&d_sm_21, gridSize2*sizeof(int)));
    gpuErrchk(cudaMalloc((void**)&d_sm_22, gridSize2*sizeof(int)));

    // --- Timing individual kernels
    float time;
    cudaEvent_t start, stop;
    cudaEventCreate(&start);
    cudaEventCreate(&stop);
    cudaEventRecord(start, 0);

    for (int i=0; i<rep_num; i++) kernel_1<<<gridSize1, blockSize1>>>(d_data1, value, d_sm_11, N);   

    cudaEventRecord(stop, 0);
    cudaEventSynchronize(stop);
    cudaEventElapsedTime(&time, start, stop);
    printf("Kernel 1 - elapsed time:  %3.3f ms \n", time/rep_num);

    cudaEventRecord(start, 0);

    for (int i=0; i<rep_num; i++) kernel_2<<<gridSize2, blockSize2>>>(d_data1, value, d_sm_21, N);   

    cudaEventRecord(stop, 0);
    cudaEventSynchronize(stop);
    cudaEventElapsedTime(&time, start, stop);
    printf("Kernel 2 - elapsed time:  %3.3f ms \n", time/rep_num);

    // --- No stream case
    cudaEventRecord(start, 0);

    kernel_1<<<gridSize1, blockSize1>>>(d_data1, value, d_sm_11, N);   
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
    gpuErrchk(cudaMemcpy(h_data1, d_data1, N*sizeof(float), cudaMemcpyDeviceToHost));
    // --- Results check
    for (int i=0; i<N; i++) {
        if (h_data1[i] != h_data1_ref[i]) {
            printf("Kernel1 - Error at i = %i; Host = %f; Device = %f\n", i, h_data1_ref[i], h_data1[i]);
            return;
        }
    }
#endif
    kernel_2<<<gridSize2, blockSize2>>>(d_data1, value, d_sm_21, N);   
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
#endif
    kernel_1<<<gridSize1, blockSize1>>>(d_data2, value, d_sm_12, N); 
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
    gpuErrchk(cudaMemcpy(d_data2, h_data2, N*sizeof(float), cudaMemcpyHostToDevice));
#endif
    kernel_2<<<gridSize2, blockSize2>>>(d_data2, value, d_sm_22, N); 
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
    gpuErrchk(cudaMemcpy(h_data2, d_data2, N*sizeof(float), cudaMemcpyDeviceToHost));
    for (int i=0; i<N; i++) {
        if (h_data2[i] != h_data2_ref[i]) {
            printf("Kernel2 - Error at i = %i; Host = %f; Device = %f\n", i, h_data2_ref[i], h_data2[i]);
            return;
        }
    }
#endif

    cudaEventRecord(stop, 0);
    cudaEventSynchronize(stop);
    cudaEventElapsedTime(&time, start, stop);
    printf("No stream - elapsed time:  %3.3f ms \n", time);

    // --- Stream case
    cudaEventRecord(start, 0);

    kernel_1<<<gridSize1, blockSize1, 0, stream1 >>>(d_data1, value, d_sm_11, N); 
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
#endif
    kernel_1<<<gridSize1, blockSize1, 0, stream2 >>>(d_data2, value, d_sm_12, N); 
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
#endif
    kernel_2<<<gridSize2, blockSize2, 0, stream1 >>>(d_data1, value, d_sm_21, N); 
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
#endif
    kernel_2<<<gridSize2, blockSize2, 0, stream2 >>>(d_data2, value, d_sm_22, N); 
#ifdef DEBUG_MODE
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize()); 
#endif

    cudaEventRecord(stop, 0);
    cudaEventSynchronize(stop);
    cudaEventElapsedTime(&time, start, stop);
    printf("Stream - elapsed time:  %3.3f ms \n", time);

    cudaStreamDestroy(stream1); 
    cudaStreamDestroy(stream2); 

    printf("Test passed!\n");

    gpuErrchk(cudaMemcpy(h_sm_11, d_sm_11, gridSize1*sizeof(int), cudaMemcpyDeviceToHost));
    gpuErrchk(cudaMemcpy(h_sm_12, d_sm_12, gridSize1*sizeof(int), cudaMemcpyDeviceToHost));
    gpuErrchk(cudaMemcpy(h_sm_21, d_sm_21, gridSize2*sizeof(int), cudaMemcpyDeviceToHost));
    gpuErrchk(cudaMemcpy(h_sm_22, d_sm_22, gridSize2*sizeof(int), cudaMemcpyDeviceToHost));

    printf("Kernel 1: gridSize = %i; blockSize = %i\n", gridSize1, blockSize1);
    printf("Kernel 2: gridSize = %i; blockSize = %i\n", gridSize2, blockSize2);
    for (int i=0; i<gridSize1; i++) {
        printf("Kernel 1 - Data 1: blockNumber = %i; SMID = %d\n", i, h_sm_11[i]);
        printf("Kernel 1 - Data 2: blockNumber = %i; SMID = %d\n", i, h_sm_12[i]);
    }
    for (int i=0; i<gridSize2; i++) {
        printf("Kernel 2 - Data 1: blockNumber = %i; SMID = %d\n", i, h_sm_21[i]);
        printf("Kernel 2 - Data 2: blockNumber = %i; SMID = %d\n", i, h_sm_22[i]);
    }
    cudaDeviceReset();

    return 0;
}

内核时序N = 100和N = 10000

N = 100
kernel_1    0.003ms
kernel_2    0.005ms    

N = 10000
kernel_1    0.011ms
kernel_2    0.053ms

因此，内核 1 在计算上比内核 2 更昂贵。

结果N = 100

Kernel 1: gridSize = 1; blockSize = 100
Kernel 2: gridSize = 1; blockSize = 100
Kernel 1 - Data 1: blockNumber = 0; SMID = 0
Kernel 1 - Data 2: blockNumber = 0; SMID = 1
Kernel 2 - Data 1: blockNumber = 0; SMID = 0
Kernel 2 - Data 2: blockNumber = 0; SMID = 1

在这种情况下，每个内核只用一个块启动，这就是时间线。

在此处输入图像描述

如您所见，发生了重叠。通过查看上述结果，调度程序将两个调用的单个块并行传递到内核 1 到两个可用的 SM，然后对内核 2 执行相同的操作。这似乎是发生重叠的主要原因。

结果N = 10000

Kernel 1: gridSize = 14; blockSize = 768
Kernel 2: gridSize = 10; blockSize = 1024
Kernel 1 - Data 1: blockNumber = 0; SMID = 0
Kernel 1 - Data 2: blockNumber = 0; SMID = 1
Kernel 1 - Data 1: blockNumber = 1; SMID = 1
Kernel 1 - Data 2: blockNumber = 1; SMID = 0
Kernel 1 - Data 1: blockNumber = 2; SMID = 0
Kernel 1 - Data 2: blockNumber = 2; SMID = 1
Kernel 1 - Data 1: blockNumber = 3; SMID = 1
Kernel 1 - Data 2: blockNumber = 3; SMID = 0
Kernel 1 - Data 1: blockNumber = 4; SMID = 0
Kernel 1 - Data 2: blockNumber = 4; SMID = 1
Kernel 1 - Data 1: blockNumber = 5; SMID = 1
Kernel 1 - Data 2: blockNumber = 5; SMID = 0
Kernel 1 - Data 1: blockNumber = 6; SMID = 0
Kernel 1 - Data 2: blockNumber = 6; SMID = 0
Kernel 1 - Data 1: blockNumber = 7; SMID = 1
Kernel 1 - Data 2: blockNumber = 7; SMID = 1
Kernel 1 - Data 1: blockNumber = 8; SMID = 0
Kernel 1 - Data 2: blockNumber = 8; SMID = 1
Kernel 1 - Data 1: blockNumber = 9; SMID = 1
Kernel 1 - Data 2: blockNumber = 9; SMID = 0
Kernel 1 - Data 1: blockNumber = 10; SMID = 0
Kernel 1 - Data 2: blockNumber = 10; SMID = 0
Kernel 1 - Data 1: blockNumber = 11; SMID = 1
Kernel 1 - Data 2: blockNumber = 11; SMID = 1
Kernel 1 - Data 1: blockNumber = 12; SMID = 0
Kernel 1 - Data 2: blockNumber = 12; SMID = 1
Kernel 1 - Data 1: blockNumber = 13; SMID = 1
Kernel 1 - Data 2: blockNumber = 13; SMID = 0
Kernel 2 - Data 1: blockNumber = 0; SMID = 0
Kernel 2 - Data 2: blockNumber = 0; SMID = 0
Kernel 2 - Data 1: blockNumber = 1; SMID = 1
Kernel 2 - Data 2: blockNumber = 1; SMID = 1
Kernel 2 - Data 1: blockNumber = 2; SMID = 1
Kernel 2 - Data 2: blockNumber = 2; SMID = 0
Kernel 2 - Data 1: blockNumber = 3; SMID = 0
Kernel 2 - Data 2: blockNumber = 3; SMID = 1
Kernel 2 - Data 1: blockNumber = 4; SMID = 1
Kernel 2 - Data 2: blockNumber = 4; SMID = 0
Kernel 2 - Data 1: blockNumber = 5; SMID = 0
Kernel 2 - Data 2: blockNumber = 5; SMID = 1
Kernel 2 - Data 1: blockNumber = 6; SMID = 1
Kernel 2 - Data 2: blockNumber = 6; SMID = 0
Kernel 2 - Data 1: blockNumber = 7; SMID = 0
Kernel 2 - Data 2: blockNumber = 7; SMID = 1
Kernel 2 - Data 1: blockNumber = 8; SMID = 1
Kernel 2 - Data 2: blockNumber = 8; SMID = 0
Kernel 2 - Data 1: blockNumber = 9; SMID = 0
Kernel 2 - Data 2: blockNumber = 9; SMID = 1

这是时间线：

在此处输入图像描述

在这种情况下，不会发生重叠。根据上述结果，这并不意味着两个 SM 没有同时被利用，而是（我认为）由于要启动的块数量较多，分配两个不同内核的块或相同的两个块内核在性能方面没有太大区别，因此调度程序选择第二个选项。

我已经测试过，考虑到每个线程完成的工作更多，行为保持不变。

于 2014-08-01T07:42:23.420 回答