从 CUDA 中的 threadId 获取 int

Question

我对 CUDA 很陌生。我需要在计算中使用线程 ID，但它不起作用。 rem 始终为 0。我需要线程的索引来计算数组中的索引，因此我无法将它们转换为浮点数来进行计算。

内核如下：

_global__ void initializationCubes(float* dVer, float* dCub, int gridSize, float* test)
{   
    int index=blockIdx.x*blockDim.x+threadIdx.x;

    if(index<(gridSize*gridSize*gridSize))
    {

        // conversion index -> i,j,k

        int rem=index;
        int qot=(rem/gridSize);

        int i=rem-(qot*gridSize);

        rem=(rem)/(gridSize);
        qot=(rem/gridSize);

        int j=rem-(qot*gridSize);

        rem=(rem)/(gridSize);
        qot=(rem/gridSize);

        int k=rem-(qot*gridSize);

            for(int x=0;x<7;x++){

             // these first three are used to test
              dCub[index*56+0+x] =index;
              dCub[index*56+7+x] =rem;
              dCub[index*56+14+x]=k;
              dCub[index*56+21+x]=dVer[((i*(gridSize+1)+(j+1))*(gridSize+1)+k)*7+x];
              dCub[index*56+28+x]=dVer[(((i+1)*(gridSize+1)+(j))*(gridSize+1)+k)*7+x];
              dCub[index*56+35+x]=dVer[(((i+1)*(gridSize+1)+(j))*(gridSize+1)+k+1)*7+x];
              dCub[index*56+42+x]=dVer[(((i+1)*(gridSize+1)+(j+1))*(gridSize+1)+k+1)*7+x];
              dCub[index*56+49+x]=dVer[(((i+1)*(gridSize+1)+(j+1))*(gridSize+1)+k)*7+x];

             }

    }   

}


__global__ void initializationVertices(float* dVer, int gridSize){


   int currentVertex=0;

   for(int i=0; i<gridSize+1; i++)
   {
       for(int j=0; j<gridSize+1; j++)
       {
          for(int k=0; k<gridSize+1; k++)
          {

               dVer[currentVertex+0]=((i*2.0f)/(gridSize)-1.0f)*2.0f;
               dVer[currentVertex+1]=((j*2.0f)/(gridSize)-1.0f)*2.0f;
               dVer[currentVertex+2]=((k*2.0f)/(gridSize)-1.0f)*2.0f;

               currentVertex+=7;
          }
       }
 }



extern "C"
void initializationCUDA1( const int verticesAtEndsOfEdges[24], const int eTable[256], int gSize, int numberParticles ) {

 numParticles=numberParticles;

 gridSize=gSize;

 numVertices=(gridSize+1)*(gridSize+1)*(gridSize+1);
 numCubes=(gridSize)*(gridSize)*(gridSize);

 size_t pitchv=7;
 cudaMallocPitch((void**)&dVer, &pitchv, 7 * sizeof(float), (gridSize+1)*(gridSize+1)*(gridSize+1));

 size_t pitchc=7;
 cudaMallocPitch((void**)&dCub, &pitchc, 7 * sizeof(float), (gridSize)*(gridSize)*(gridSize)*8);

 cudaMalloc((void **)&verticesAtEnds, 24*sizeof(int));

 cudaMalloc((void **)&dedgeTable, 256*sizeof(int));

 cudaMalloc((void **)&dtriTable, 256*16*sizeof(int));

 cudaMalloc((void **)&ballPoint, 3*sizeof(float));

 cudaMalloc((void **)&dpositions, 3*numberParticles*sizeof(float));

 cudaMalloc((void **)&dedgeVertices, numCubes*6*12*sizeof(float));

 cudaMalloc((void **)&result, numCubes*18*sizeof(float));

 output=(float*)malloc(numCubes*18*sizeof(float));

 cudaMalloc((void **)&numFaces, 10*sizeof(int));

 cudaMalloc((void **)&test, sizeof(float));




 initializationVertices<<<1,1>>>(dVer, gridSize);

 initializationCubes<<<128,256>>>( dVer, dCub, gridSize, test);

 float* tmp =(float*)malloc(numCubes*56*(sizeof(float)));

 cudaMemcpy(tmp, dCub, numCubes*56*sizeof(float), cudaMemcpyDeviceToHost);
 for(int a=0;a<100;a++){
   printf("%f\n",tmp[a]);
 }
}

编辑

gridSize 是 40 -> 线程的迭代从 0 到 64000

当我打印函数之外的值时，rem、i、j 和 k 都等于 0。

size_t pitchv=7; cudaMallocPitch((void**)&dVer, &pitchv, 7 * sizeof(float), (gridSize+1)(gridSize+1)(gridSize+1));

size_t pitchc=7; cudaMallocPitch((void**)&dCub, &pitchc, 7 * sizeof(float), (gridSize)(gridSize)(gridSize)*8);

initializationCubes>>(dVer, dCub, gridSize, test);

Answer 1

如果gridSize 是网格的大小，顾名思义，rem 和qot 在执行代码后都将始终为零，因为它们除以一个比它们自己更大的值。

如果您正在寻找三维网格的索引，这正是threadIdx 和blockIdx 具有三个组件的原因。根本不需要昂贵的除法，只需使用这个标准代码sn-p：

int i = blockIdx.x * blockDim.x + threadIdx.x;
int j = blockIdx.y * blockDim.y + threadIdx.y;
int k = blockIdx.z * blockDim.z + threadIdx.z;

if (i < myBlockSize.x && j < myBlockSize.y && k<myBlockSize.z) {
    // your kernel code...
}

并使用块和网格大小的 y 和 z 分量的适当值以及设置为所需网格大小的参数或全局变量 myBlockSize 启动内核（以防它不能被分解为整数块-和网格尺寸）。