在编写一些CUDA代码时遇到了这个非常奇怪的问题:从gpu到cpu内存的同一块cudaMemcpy需要不同的时间来完成对子程序的不同迭代调用,这是一个巨大的差异:~60 ms vs~0.02 ms。
代码如下:
float calc_formation_obj( int formationNo, bool calcObj )
{
int i;
int prev = prevCP[aperIndex];
int next = nextCP[aperIndex];
float ll = formations_l[formationNo];
float rl = formations_r[formationNo];
float obj = 0.0;
float *f_grid = new float[grid_size_voxe];
// use ll and rl
thrust::device_ptr<float> dll(d_leafpos_l);
thrust::device_ptr<float> drl(d_leafpos_r);
dll[rows_per_beam*aperIndex+ rowIndex] = ll;
drl[rows_per_beam*aperIndex+ rowIndex] = rl;
// set all leaf positions between prev/next
set_leafpos<<<grid_size_ncps,BLOCK_SIZE>>> (aperIndex, rowIndex, prev, next, ncps, d_leafpos_l,
d_leafpos_r, ll, rl, rows_per_beam, d_cp_angles);
// copy dose to dose_temp
thrust::device_ptr<float> ddose(d_dose);
thrust::device_ptr<float> dtp(d_dose_temp);
thrust::copy(ddose, ddose+nvoxel, dtp);
// the angles actually being added
if (prev==-1) {
prev = 0;
}
if (next==ncps) {
next = ncps-1;
}
// add dose from all these leaf positions
// if last arg 1 then add
add_remove_dose<<<grid_size_ncps,BLOCK_SIZE>>> (prev,next, rowIndex, d_dose_temp, d_leafpos_l,
d_leafpos_r, d_voxe_b, d_dijs_b, d_voxnumperbixcum, d_flu_cp, rows_per_beam, bix_per_row, beamletSize, 1);
if (!calcObj) {
return(0.0);
}
// initialize
cudaMemset((void*)d_f_voxel, 0, voxesize_f);
cudaMemset((void*)d_f_grid, 0, sizeof(float)*grid_size_voxe);
// then calculate objective
calc_obj_dose<<<grid_size_voxe,BLOCK_SIZE>>>( d_dose_temp, d_f_voxel, d_thresh, d_is_target, nvoxel,
d_f_grid, d_od_wt, d_ud_wt );
// copy results from GPU
time_t time_1,time_2;
float elapse;
time_1=clock();
cudaMemcpy(f_grid, d_f_grid, sizeof(float)*grid_size_voxe, cudaMemcpyDeviceToHost);
time_2 = clock();
elapse = ((float)time_2 - (float)time_1)/1000;
printf( "iter %d copy time: %f ms\n", formationNo, elapse );
obj = 0.0;
for (i=0; i<grid_size_voxe; i++) {
obj += f_grid[i];
}
delete[] f_grid;
return(obj);
}
在程序中多次调用此子例程,每次运行时都会记录
的运行时cudaMemcpy(f_grid, d_f_grid, sizeof(float)*grid_size_voxe, cudaMemcpyDeviceToHost);
我得到的结果如下:
iter 0 copy time: 0.018000 ms
iter 1 copy time: 66.445999 ms
iter 2 copy time: 64.239998 ms
iter 3 copy time: 66.959999 ms
iter 4 copy time: 66.328003 ms
iter 5 copy time: 65.656998 ms
iter 6 copy time: 66.120003 ms
iter 7 copy time: 63.811001 ms
iter 8 copy time: 66.530998 ms
iter 9 copy time: 65.686996 ms
iter 10 copy time: 65.808998 ms
iter 11 copy time: 0.027000 ms
iter 12 copy time: 64.346001 ms
iter 13 copy time: 66.407997 ms
iter 14 copy time: 65.796997 ms
iter 15 copy time: 65.471001 ms
iter 16 copy time: 66.209000 ms
iter 17 copy time: 63.799000 ms
iter 18 copy time: 66.542999 ms
iter 19 copy time: 65.660004 ms
iter 20 copy time: 65.102997 ms
iter 21 copy time: 0.019000 ms
iter 22 copy time: 64.665001 ms
iter 23 copy time: 66.653999 ms
iter 24 copy time: 65.648003 ms
iter 25 copy time: 65.233002 ms
iter 26 copy time: 65.851997 ms
iter 27 copy time: 63.992001 ms
iter 28 copy time: 66.172997 ms
iter 29 copy time: 65.503998 ms
iter 30 copy time: 0.020000 ms
iter 31 copy time: 66.277000 ms
iter 32 copy time: 63.881001 ms
iter 33 copy time: 66.537003 ms
iter 34 copy time: 65.626999 ms
iter 35 copy time: 65.387001 ms
iter 36 copy time: 66.084999 ms
iter 37 copy time: 63.797001 ms
iter 38 copy time: 0.017000 ms
iter 39 copy time: 65.707001 ms
iter 40 copy time: 65.553001 ms
iter 41 copy time: 66.362999 ms
iter 42 copy time: 63.634998 ms
使用CUDA 4.2在带有GeForce GTX 285的Mac OSX上运行。我不知道为什么这样做 - 应该是一个简单的副本。任何帮助表示赞赏!!
答案 0 :(得分:5)
cudaMemcpy
阻塞,直到上一个内核完成。你确定你没有测量内核性能而不是复制性能吗?在开始计时cudaDeviceSynchronize()
之前,请投入cudaMemcpy
。
答案 1 :(得分:0)
不应使用clock()来测量时间,而应使用事件:
使用事件你会有这样的事情:
cudaEvent_t start, stop; // variables that holds 2 events
float time; // Variable that will hold the time
cudaEventCreate(&start); // creating the event 1
cudaEventCreate(&stop); // creating the event 2
cudaEventRecord(start, 0); // start measuring the time
// What you want to measure
cudaMemcpy(f_grid, d_f_grid, sizeof(float)*grid_size_voxe, cudaMemcpyDeviceToHost);
cudaEventRecord(stop, 0); // Stop time measuring
cudaEventSynchronize(stop); // Wait until the completion of all device
// work preceding the most recent call to cudaEventRecord()
cudaEventElapsedTime(&time, start, stop); // Saving the time measured