CUDA学习笔记(三)CUDA简介

本篇博文转载于https://www.cnblogs.com/1024incn/tag/CUDA/,仅用于学习。

前言

线程的组织形式对程序的性能影响是至关重要的,本篇博文主要以下面一种情况来介绍线程组织形式:

  • 2D grid 2D block

线程索引

矩阵在memory中是row-major线性存储的:

在kernel里,线程的唯一索引非常有用,为了确定一个线程的索引,我们以2D为例:

  • 线程和block索引
  • 矩阵中元素坐标
  • 线性global memory 的偏移

首先可以将thread和block索引映射到矩阵坐标:

ix = threadIdx.x + blockIdx.x * blockDim.x

iy = threadIdx.y + blockIdx.y * blockDim.y

之后可以利用上述变量计算线性地址:

idx = iy * nx + ix

 

CUDA学习笔记(三)CUDA简介_第1张图片

上图展示了block和thread索引,矩阵坐标以及线性地址之间的关系,谨记,相邻的thread拥有连续的threadIdx.x,也就是索引为(0,0)(1,0)(2,0)(3,0)...的thread连续,而不是(0,0)(0,1)(0,2)(0,3)...连续,跟我们线代里玩矩阵的时候不一样。

现在可以验证出下面的关系:

thread_id(2,1)block_id(1,0) coordinate(6,1) global index 14 ival 14

下图显示了三者之间的关系:

 

CUDA学习笔记(三)CUDA简介_第2张图片

代码

 

int main(int argc, char **argv) {
  printf("%s Starting...\n", argv[0]);
  // set up device
  int dev = 0;
  cudaDeviceProp deviceProp;
  CHECK(cudaGetDeviceProperties(&deviceProp, dev));
  printf("Using Device %d: %s\n", dev, deviceProp.name);
  CHECK(cudaSetDevice(dev));  // set up date size of matrix
  int nx = 1<<14;
  int ny = 1<<14;
  int nxy = nx*ny;
  int nBytes = nxy * sizeof(float);
  printf("Matrix size: nx %d ny %d\n",nx, ny);
  // malloc host memory
  float *h_A, *h_B, *hostRef, *gpuRef;
  h_A = (float *)malloc(nBytes);
  h_B = (float *)malloc(nBytes);
  hostRef = (float *)malloc(nBytes);
  gpuRef = (float *)malloc(nBytes);
    // initialize data at host side
  double iStart = cpuSecond();
  initialData (h_A, nxy);
  initialData (h_B, nxy);
  double iElaps = cpuSecond() - iStart;
  memset(hostRef, 0, nBytes);
  memset(gpuRef, 0, nBytes);
  // add matrix at host side for result checks
  iStart = cpuSecond();
  sumMatrixOnHost (h_A, h_B, hostRef, nx,ny);
  iElaps = cpuSecond() - iStart;
  // malloc device global memory
  float *d_MatA, *d_MatB, *d_MatC;
  cudaMalloc((void **)&d_MatA, nBytes);
  cudaMalloc((void **)&d_MatB, nBytes);
  cudaMalloc((void **)&d_MatC, nBytes);
    // transfer data from host to device
  cudaMemcpy(d_MatA, h_A, nBytes, cudaMemcpyHostToDevice);
  cudaMemcpy(d_MatB, h_B, nBytes, cudaMemcpyHostToDevice);
  // invoke kernel at host side
  int dimx = 32;
  int dimy = 32;
  dim3 block(dimx, dimy);
  dim3 grid((nx+block.x-1)/block.x, (ny+block.y-1)/block.y);
  iStart = cpuSecond();
  sumMatrixOnGPU2D <<< grid, block >>>(d_MatA, d_MatB, d_MatC, nx, ny);
  cudaDeviceSynchronize();
  iElaps = cpuSecond() - iStart;
  printf("sumMatrixOnGPU2D <<<(%d,%d), (%d,%d)>>> elapsed %f sec\n", grid.x,
  grid.y, block.x, block.y, iElaps);
  // copy kernel result back to host side
  cudaMemcpy(gpuRef, d_MatC, nBytes, cudaMemcpyDeviceToHost);
  // check device results
  checkResult(hostRef, gpuRef, nxy);
    // free device global memory
  cudaFree(d_MatA);
  cudaFree(d_MatB);
  cudaFree(d_MatC);
  // free host memory
  free(h_A);
  free(h_B);
  free(hostRef);
  free(gpuRef);
  // reset device
  cudaDeviceReset();
  return (0);
}

编译运行:

$ nvcc -arch=sm_20 sumMatrixOnGPU-2D-grid-2D-block.cu -o matrix2D
$ ./matrix2D

输出:

./a.out Starting...
Using Device 0: Tesla M2070
Matrix size: nx 16384 ny 16384
sumMatrixOnGPU2D <<<(512,512), (32,32)>>> elapsed 0.060323 sec
Arrays match.

接下来,我们更改block配置为32x16,重新编译,输出为:

sumMatrixOnGPU2D <<<(512,1024), (32,16)>>> elapsed 0.038041 sec

可以看到,性能提升了一倍,直观的来看,我们会认为第二个配置比第一个多了一倍的block所以性能提升一倍,实际上也确实是因为block增加了。但是,如果你继续增加block的数量,则性能又会降低:

sumMatrixOnGPU2D <<< (1024,1024), (16,16) >>> elapsed 0.045535 sec

下图展示了不同配置的性能;

 

CUDA学习笔记(三)CUDA简介_第3张图片

关于性能的分析将在之后的博文中总结,现在只是了解下,本文在于掌握线程组织的方法。

 

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