concurrency - CUDA 流和并发内核执行

Question

我想使用流来并行执行在单独的设备数据阵列上工作的内核。数据在设备上分配并从以前的内核填充。

我编写了以下程序，表明我目前无法达到我的目标。事实上，两个非默认流上的内核在各自的流中顺序执行。

在 2 台装有最新 Debian Linux 版本的 Intel 机器上观察到同样的行为。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。

代码如下:

#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 512^3单例float .我通常只是将数组分成 (512,1,1) block 。我放在大小为 (1<<15, (rest), 1) 的网格上的线程.

提前感谢您的任何提示或评论。

最好的问候。

Answer 1

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 block 在哪个流式多处理器 (SM) 上运行。我使用的是 CUDA 6.5(候选版本)，我在 GT540M 卡上运行，它只有 2 SM，因此它提供了一个简单的工作场所。 blockSize 选择委托(delegate)给新的 CUDA 6.5 cudaOccupancyMaxPotentialBlockSize 工具。

代码

#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;
}

N = 100 和 N = 10000 的内核时序

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 的计算成本高于内核 2。

N = 100的结果

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，每个内核仅启动一个 block ，这是时间线。

如您所见，发生了重叠。通过查看上述结果，调度程序将两次调用的单个 block 并行地传递给两个可用的 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。

N = 10000的结果

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:

在这种情况下，不会发生重叠。根据以上结果，这并不意味着两个 SM 没有同时被利用，而是(我认为)，由于要启动的区 block 数量较多，分配两个不同内核的区 block 或相同的两个区 block 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。