CUDA流和并发内核执行
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中填充。
我写了下面的程序，显示我目前无法达到目标。 事实上，两个非默认流上的内核在它们各自的流中顺序执行。
在具有最新Debian linux版本的2台英特尔机器上观察到相同的行为。 其中一款采用CUDA 4.2的特斯拉C2075，另一款采用CUDA 5.0的Geforce 460GT。 Visual Profiler在4.2和5.0 CUDA版本中均显示顺序执行。
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 stuffn");
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，数组的维数为512^3单float 。 我通常只是将(512,1,1)线程的数组切割成一个大小的网格(1<<15, (rest), 1) 。
提前感谢您的任何提示或评论。
最好的祝福。
我试图提供一个解释，为什么你没有看到你的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视每个块运行哪个流式多处理器（SM）。 我使用的是CUDA 6.5（Release Candidate），我使用的GT540M卡只有2 SM，所以它提供了一个简单的操作环境。 blockSize选项被委托给新的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 %dn", 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 = %fn", 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 = %fn", 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 = %in", gridSize1, blockSize1);
printf("Kernel 2: gridSize = %i; blockSize = %in", gridSize2, blockSize2);
for (int i=0; i<gridSize1; i++) {
printf("Kernel 1 - Data 1: blockNumber = %i; SMID = %dn", i, h_sm_11[i]);
printf("Kernel 1 - Data 2: blockNumber = %i; SMID = %dn", i, h_sm_12[i]);
}
for (int i=0; i<gridSize2; i++) {
printf("Kernel 2 - Data 1: blockNumber = %i; SMID = %dn", i, h_sm_21[i]);
printf("Kernel 2 - Data 2: blockNumber = %i; SMID = %dn", i, h_sm_22[i]);
}
cudaDeviceReset();
return 0;
}
N = 100和N = 10000内核时间
N = 100
kernel_1 0.003ms
kernel_2 0.005ms
N = 10000
kernel_1 0.011ms
kernel_2 0.053ms
所以，内核1比内核2在计算上更昂贵。
N = 100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块，这就是时间线。
如你所见，重叠发生。 通过查看上述结果，调度程序将两个调用的单个块与两个可用的SM并行地发送到内核1，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
这是时间表：
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不会同时被利用，但是（我认为）由于要启动的块数量较多，分配两块不同的内核或两块相同的块内核在性能方面没有太大的差别，因此调度器选择了第二个选项。
我已经测试过，考虑到每个线程完成更多的工作，行为保持不变。
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