[英]CUDA Zero Copy vs. CudaMemcpy on Jetson TK1
我的問題:我正在尋找某人要么指出我試圖在CUDA中使用實現零拷貝的方式中的錯誤,要么揭示更多“幕后”視角為什么零拷貝方法不會更快比memcpy方法。 順便說一下,我正在使用Ubuntu對NVidia的TK1處理器進行測試。
我的問題與使用CIDA有效地使用NVIDIA TK1(物理)統一內存架構有關。 NVIDIA提供了兩種GPU / CPU內存傳輸抽象方法。
我的測試代碼的簡短描述:我使用方法1和2測試了相同的cuda內核。鑒於源數據的設備沒有復制或結果數據的設備沒有復制,我預計1會更快。 然而,結果倒退到我的假設(方法#1慢50%)。 以下是我測試的代碼:
#include <libfreenect/libfreenect.hpp>
#include <iostream>
#include <vector>
#include <cmath>
#include <pthread.h>
#include <cxcore.h>
#include <time.h>
#include <sys/time.h>
#include <memory.h>
///CUDA///
#include <cuda.h>
#include <cuda_runtime.h>
///OpenCV 2.4
#include <highgui.h>
#include <cv.h>
#include <opencv2/gpu/gpu.hpp>
using namespace cv;
using namespace std;
///The Test Kernel///
__global__ void cudaCalcXYZ( float *dst, float *src, float *M, int height, int width, float scaleFactor, int minDistance)
{
float nx,ny,nz, nzpminD, jFactor;
int heightCenter = height / 2;
int widthCenter = width / 2;
//int j = blockIdx.x; //Represents which row we are in
int index = blockIdx.x*width;
jFactor = (blockIdx.x - heightCenter)*scaleFactor;
for(int i= 0; i < width; i++)
{
nz = src[index];
nzpminD = nz + minDistance;
nx = (i - widthCenter )*(nzpminD)*scaleFactor;
ny = (jFactor)*(nzpminD);
//Solve for only Y matrix (height vlaues)
dst[index++] = nx*M[4] + ny*M[5] + nz*M[6];
//dst[index++] = 1 + 2 + 3;
}
}
//Function fwd declarations
double getMillis();
double getMicros();
void runCudaTestZeroCopy(int iter, int cols, int rows);
void runCudaTestDeviceCopy(int iter, int cols, int rows);
int main(int argc, char **argv) {
//ZERO COPY FLAG (allows runCudaTestZeroCopy to run without fail)
cudaSetDeviceFlags(cudaDeviceMapHost);
//Runs kernel using explicit data copy to 'device' and back from 'device'
runCudaTestDeviceCopy(20, 640,480);
//Uses 'unified memory' cuda abstraction so device can directly work from host data
runCudaTestZeroCopy(20,640, 480);
std::cout << "Stopping test" << std::endl;
return 0;
}
void runCudaTestZeroCopy(int iter, int cols, int rows)
{
cout << "CUDA Test::ZEROCOPY" << endl;
int src_rows = rows;
int src_cols = cols;
int m_rows = 4;
int m_cols = 4;
int dst_rows = src_rows;
int dst_cols = src_cols;
//Create and allocate memory for host mats pointers
float *psrcMat;
float *pmMat;
float *pdstMat;
cudaHostAlloc((void **)&psrcMat, src_rows*src_cols*sizeof(float), cudaHostAllocMapped);
cudaHostAlloc((void **)&pmMat, m_rows*m_cols*sizeof(float), cudaHostAllocMapped);
cudaHostAlloc((void **)&pdstMat, dst_rows*dst_cols*sizeof(float), cudaHostAllocMapped);
//Create mats using host pointers
Mat src_mat = Mat(cvSize(src_cols, src_rows), CV_32FC1, psrcMat);
Mat m_mat = Mat(cvSize(m_cols, m_rows), CV_32FC1, pmMat);
Mat dst_mat = Mat(cvSize(dst_cols, dst_rows), CV_32FC1, pdstMat);
//configure src and m mats
for(int i = 0; i < src_rows*src_cols; i++)
{
psrcMat[i] = (float)i;
}
for(int i = 0; i < m_rows*m_cols; i++)
{
pmMat[i] = 0.1234;
}
//Create pointers to dev mats
float *d_psrcMat;
float *d_pmMat;
float *d_pdstMat;
//Map device to host pointers
cudaHostGetDevicePointer((void **)&d_psrcMat, (void *)psrcMat, 0);
//cudaHostGetDevicePointer((void **)&d_pmMat, (void *)pmMat, 0);
cudaHostGetDevicePointer((void **)&d_pdstMat, (void *)pdstMat, 0);
//Copy matrix M to device
cudaMalloc( (void **)&d_pmMat, sizeof(float)*4*4 ); //4x4 matrix
cudaMemcpy( d_pmMat, pmMat, sizeof(float)*m_rows*m_cols, cudaMemcpyHostToDevice);
//Additional Variables for kernels
float scaleFactor = 0.0021;
int minDistance = -10;
//Run kernel! //cudaSimpleMult( float *dst, float *src, float *M, int width, int height)
int blocks = src_rows;
const int numTests = iter;
double perfStart = getMillis();
for(int i = 0; i < numTests; i++)
{
//cudaSimpleMult<<<blocks,1>>>(d_pdstMat, d_psrcMat, d_pmMat, src_cols, src_rows);
cudaCalcXYZ<<<blocks,1>>>(d_pdstMat, d_psrcMat, d_pmMat, src_rows, src_cols, scaleFactor, minDistance);
cudaDeviceSynchronize();
}
double perfStop = getMillis();
double perfDelta = perfStop - perfStart;
cout << "Ran " << numTests << " iterations totaling " << perfDelta << "ms" << endl;
cout << " Average time per iteration: " << (perfDelta/(float)numTests) << "ms" << endl;
//Copy result back to host
//cudaMemcpy(pdstMat, d_pdstMat, sizeof(float)*src_rows*src_cols, cudaMemcpyDeviceToHost);
//cout << "Printing results" << endl;
//for(int i = 0; i < 16*16; i++)
//{
// cout << "src[" << i << "]= " << psrcMat[i] << " dst[" << i << "]= " << pdstMat[i] << endl;
//}
cudaFree(d_psrcMat);
cudaFree(d_pmMat);
cudaFree(d_pdstMat);
cudaFreeHost(psrcMat);
cudaFreeHost(pmMat);
cudaFreeHost(pdstMat);
}
void runCudaTestDeviceCopy(int iter, int cols, int rows)
{
cout << "CUDA Test::DEVICE COPY" << endl;
int src_rows = rows;
int src_cols = cols;
int m_rows = 4;
int m_cols = 4;
int dst_rows = src_rows;
int dst_cols = src_cols;
//Create and allocate memory for host mats pointers
float *psrcMat;
float *pmMat;
float *pdstMat;
cudaHostAlloc((void **)&psrcMat, src_rows*src_cols*sizeof(float), cudaHostAllocMapped);
cudaHostAlloc((void **)&pmMat, m_rows*m_cols*sizeof(float), cudaHostAllocMapped);
cudaHostAlloc((void **)&pdstMat, dst_rows*dst_cols*sizeof(float), cudaHostAllocMapped);
//Create pointers to dev mats
float *d_psrcMat;
float *d_pmMat;
float *d_pdstMat;
cudaMalloc( (void **)&d_psrcMat, sizeof(float)*src_rows*src_cols );
cudaMalloc( (void **)&d_pdstMat, sizeof(float)*src_rows*src_cols );
cudaMalloc( (void **)&d_pmMat, sizeof(float)*4*4 ); //4x4 matrix
//Create mats using host pointers
Mat src_mat = Mat(cvSize(src_cols, src_rows), CV_32FC1, psrcMat);
Mat m_mat = Mat(cvSize(m_cols, m_rows), CV_32FC1, pmMat);
Mat dst_mat = Mat(cvSize(dst_cols, dst_rows), CV_32FC1, pdstMat);
//configure src and m mats
for(int i = 0; i < src_rows*src_cols; i++)
{
psrcMat[i] = (float)i;
}
for(int i = 0; i < m_rows*m_cols; i++)
{
pmMat[i] = 0.1234;
}
//Additional Variables for kernels
float scaleFactor = 0.0021;
int minDistance = -10;
//Run kernel! //cudaSimpleMult( float *dst, float *src, float *M, int width, int height)
int blocks = src_rows;
double perfStart = getMillis();
for(int i = 0; i < iter; i++)
{
//Copty from host to device
cudaMemcpy( d_psrcMat, psrcMat, sizeof(float)*src_rows*src_cols, cudaMemcpyHostToDevice);
cudaMemcpy( d_pmMat, pmMat, sizeof(float)*m_rows*m_cols, cudaMemcpyHostToDevice);
//Run Kernel
//cudaSimpleMult<<<blocks,1>>>(d_pdstMat, d_psrcMat, d_pmMat, src_cols, src_rows);
cudaCalcXYZ<<<blocks,1>>>(d_pdstMat, d_psrcMat, d_pmMat, src_rows, src_cols, scaleFactor, minDistance);
//Copy from device to host
cudaMemcpy( pdstMat, d_pdstMat, sizeof(float)*src_rows*src_cols, cudaMemcpyDeviceToHost);
}
double perfStop = getMillis();
double perfDelta = perfStop - perfStart;
cout << "Ran " << iter << " iterations totaling " << perfDelta << "ms" << endl;
cout << " Average time per iteration: " << (perfDelta/(float)iter) << "ms" << endl;
cudaFree(d_psrcMat);
cudaFree(d_pmMat);
cudaFree(d_pdstMat);
cudaFreeHost(psrcMat);
cudaFreeHost(pmMat);
cudaFreeHost(pdstMat);
}
//Timing functions for performance measurements
double getMicros()
{
timespec ts;
//double t_ns, t_s;
long t_ns;
double t_s;
clock_gettime(CLOCK_MONOTONIC, &ts);
t_s = (double)ts.tv_sec;
t_ns = ts.tv_nsec;
//return( (t_s *1000.0 * 1000.0) + (double)(t_ns / 1000.0) );
return ((double)t_ns / 1000.0);
}
double getMillis()
{
timespec ts;
double t_ns, t_s;
clock_gettime(CLOCK_MONOTONIC, &ts);
t_s = (double)ts.tv_sec;
t_ns = (double)ts.tv_nsec;
return( (t_s * 1000.0) + (t_ns / 1000000.0) );
}
我已經看過后Cuda零拷貝性能 ,但我覺得這與以下原因無關:GPU和CPU具有物理統一的內存架構。
謝謝
當您使用ZeroCopy時,對內存的讀取會經過一些路徑,在該路徑中,它會查詢內存單元以從系統內存中獲取數據。 此操作有一些延遲。
當使用直接訪問存儲器時,存儲器單元從全局存儲器收集數據,並具有不同的訪問模式和延遲。
實際上看到這種差異需要某種形式的分析。
盡管如此,您對全局函數的調用使用了單個線程
cudaCalcXYZ<<< blocks,1 >>> (...
在這種情況下,當從系統內存(或全局內存)收集內存時,GPU幾乎無法隱藏延遲。 我建議你使用更多的線程(64的一些,總共至少128),並在其上運行探查器以獲得內存訪問的成本。 您的算法似乎是可分離的,並修改了代碼
for(int i= 0; i < width; i++)
至
for (int i = threadIdx.x ; i < width ; i += blockDim.x)
可能會提高整體表現。 圖像大小為640,將變為128個線程的5次迭代。
cudaCalcXYZ<<< blocks,128 >>> (...
我相信這會帶來一些性能提升。
ZeroCopy功能允許我們在設備上運行數據,而無需手動將其復制到設備內存,如cudaMemcpy功能。 零拷貝存儲器僅將主機地址傳遞給在內核設備上讀/寫的設備。 因此,您向內核設備聲明的線程塊越多,在內核設備上讀取/寫入的數據越多,傳遞給設備的主機地址就越多。 最后,與僅向設備內核聲明一些線程塊相比,您獲得了更好的性能提升。
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