迭代之间的 Cudamemcpy 设备>主机时间更改

Question

我在编写一些 CUDA 代码时遇到了这个非常奇怪的问题：从 GPU 到 CPU 内存的同一段 cudaMemcpy 在对子程序的不同迭代调用中需要不同的时间才能完成，这是一个巨大的差异：~60 毫秒与 ~0.02 毫秒。

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 msn", formationNo, elapse );
    obj = 0.0;
    for (i=0; i<grid_size_voxe; i++) {
        obj += f_grid[i];
    }

    delete[] f_grid;
    return(obj);
}

这个子例程在程序过程中被调用了很多次，每次它运行时我都会记录运行时

cudaMemcpy(f_grid, d_f_grid, sizeof(float)*grid_size_voxe, cudaMemcpyDeviceToHost);

我得到的结果看起来像：

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

这是在带有GeForce GTX 285的Mac OSX上运行的，使用CUDA 4.2。我不知道它为什么要这样做---应该是一个简单的副本。任何帮助不胜感激！！

Answer 1

cudaMemcpy 阻塞

，直到上一个内核完成。 您确定不是在测量内核性能而不是复制性能吗？ 在开始计时cudaMemcpy之前先投入cudaDeviceSynchronize()。

Answer 2

与其使用 clock（） 来测量时间，不如使用 事件：

使用事件，你会有这样的东西：

  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