用于内环的CUDA螺纹

我有这个内核

__global__ void kernel1(int keep, int include, int width, int* d_Xco, 
              int* d_Xnum, bool* d_Xvalid, float* d_Xblas)
{

  int i, k;  
  i = threadIdx.x + blockIdx.x * blockDim.x;

  if(i < keep){

    for(k = 0; k < include ; k++){

      int val = (d_Xblas[i*include + k] >= 1e5);
      int aux = d_Xnum[i];

      d_Xblas[i*include + k] *= (!val);
      d_Xco[i*width + aux] = k;
      d_Xnum[i] +=val;
      d_Xvalid[i*include + k] = (!val);
    }
  }
}

使用二维栅格启动：

int keep = 9000;
int include = 23000;
int width = 0.2*include;

int th = 32;
dim3 threads(th,th);
dim3 blocks ((keep+threads.x-1)/threads.x, (include+threads.y-1)/threads.y);
kernel2 <<< blocks,threads >>>( keep, include, width, d_Xco, d_Xnum, 
                               d_Xvalid, d_Xblas );

---

在原始代码中，您有

for(k = 0; k < include ; k++){
  ...
  int aux = d_Xnum[i];
  ...
  d_Xco[i*width + aux] = k;
  ...
}

---

一旦您这样做，当您并行化

k

循环时，您将不再具有当前的竞争条件，因为许多

k

线程同时将不同的值分配到

dxco

中的同一位置（不保证排序）。

网格应创建为

dim3块（（keep+threads.x-1）/螺纹.x，（包括+螺纹.y-1）/threads.y）。你忘了括号了。@sgar91对不起，这是个打字错误。为什么内核2不能工作？它给出了错误的结果？它根本不执行？@pQB是的，错误的结果。它确实加快了速度，但…执行很好、快速且平稳。您的数据是如何安排的？根据内核2，您有
保留
列和
包含
行。@harrism:我确实需要
dxco
的数据，而不仅仅是最后一个元素。@Manolete，在您的顺序版本中，只有最后一个元素会保留！逐步完成调试器中的代码，请参见。
int keep = 9000;
int include = 23000;
int width = 0.2*include;

int th = 32;
dim3 threads(th,th);
dim3 blocks ((keep+threads.x-1)/threads.x, (include+threads.y-1)/threads.y);
kernel2 <<< blocks,threads >>>( keep, include, width, d_Xco, d_Xnum, 
                               d_Xvalid, d_Xblas );

d_Xco
-------------------------------
| 2|3 |15 |4 |5 |5 | | | | | | .......
-------------------------------

for(k = 0; k < include ; k++){
  ...
  int aux = d_Xnum[i];
  ...
  d_Xco[i*width + aux] = k;
  ...
}

 d_Xco[i*width + d_Xnum[i]] = include - 1;