Question

我传输48kb数据块（带固定内存），虽然cuda事件看到它以5gb /秒的速度上升，但当我们回到窗口时，我们只能看到一半的速度。这只是不可避免的驱动程序开销，还是有办法缓解这种情况？我已将该过程封装在下面的测试程序中。

void transferUp(size_t size)
{

StopWatchWin timer;
timer.start();

float tUpCopyStart,tUpCopyStop;

cudaEvent_t sendUpStopEvent,sendUpStartEvent;
checkCudaErrors(cudaEventCreate( &sendUpStartEvent ));
checkCudaErrors(cudaEventCreate( &sendUpStopEvent ));

unsigned *cpu_sending = (unsigned *)malloc(size);
checkCudaErrors(cudaHostAlloc(&cpu_sending, size*sizeof(unsigned), cudaHostAllocPortable));

unsigned *gpu_receiving;
checkCudaErrors(cudaMalloc(&gpu_receiving, size*sizeof(unsigned)));

tUpCopyStart = timer.getTime();
checkCudaErrors(cudaEventRecord(sendUpStartEvent));

checkCudaErrors(cudaMemcpyAsync(gpu_receiving, cpu_sending, size*sizeof(unsigned), cudaMemcpyHostToDevice));

checkCudaErrors(cudaEventRecord(sendUpStopEvent));
checkCudaErrors(cudaEventSynchronize(sendUpStopEvent));
tUpCopyStop = timer.getTime();

double sendTimeWindows = tUpCopyStop - tUpCopyStart;

float sendTimeCuda;
checkCudaErrors(cudaEventElapsedTime( &sendTimeCuda,sendUpStartEvent,sendUpStopEvent));

float GbSec_cuda = (size*sizeof(unsigned)/1000)/(sendTimeCuda*1000);
float GbSec_win = (size*sizeof(unsigned)/1000)/(sendTimeWindows*1000);

printf("size=%06d bytes eventTime=%.03fms windowsTime=%0.3fms cudaSpeed=%.01f gb/s winSpeed=%.01f gb/s\n",
size*sizeof(unsigned),sendTimeCuda,sendTimeWindows,GbSec_cuda,GbSec_win);

checkCudaErrors(cudaEventDestroy( sendUpStartEvent ));
checkCudaErrors(cudaEventDestroy( sendUpStopEvent ));

checkCudaErrors(cudaFreeHost(cpu_sending));
checkCudaErrors(cudaFree(gpu_receiving));

}

Answer 1

对这个小操作进行计时的开销是压倒性的。

对于小型主机 - >设备副本（例如，64K或更小），CUDA驱动程序会将数据内联到命令缓冲区中，因此即使是声称同步的memcpy调用实际上也是异步完成的。但是，代码中的cudaEventSynchronize()调用会强制CPU等待而不是继续执行。

使用事件与窗口的CUDA传输时间

1 个答案: