在CUDA中汇总矩阵的行（以行优先或列优先顺序存储） [英] Summing the rows of a matrix (stored in either row-major or column-major order) in CUDA

问题描述

我正在研究将CUDA中矩阵的行相加的问题。我给出以下示例。

I'm working on the problem summing the rows of a matrix in CUDA. I'm giving the following example.

假设具有以下 20 * 4 数组：

1 2 3 4
4 1 2 3
3 4 1 2 
.
1 2 3 4
.
.
.
.
.
.
.
.
2 1 3 4

将2d数组展平为1d数组后-major或column-major顺序），我需要将每个线程分配到不同的行并计算该行的成本。

After flattened the 2d array to a 1d array (either in row-major or column-major order), I need to assign each thread to a different row and calculate the cost for that row.

例如

-线程1应该计算 1 2 3 4

-线程2应该计算 4 1 2 3

如何在CUDA中做到这一点？

谢谢大家的答复

推荐答案

#include <stdio.h>
#include <stdlib.h>
#define MROWS 20
#define NCOLS 4
#define nTPB 256

__global__ void mykernel(int *costdata, int rows, int cols, int *results){
  int tidx = threadIdx.x + blockDim.x*blockIdx.x;
  if (tidx < rows){
    int mycost = 0;
    for (int i = 0; i < cols; i++)
       mycost += costdata[(tidx*cols)+i];
    results[tidx] = mycost;
    }
  }

int main(){
  //define and initialize host and device storage for cost and results
  int *d_costdata, *h_costdata, *d_results, *h_results;
  h_results = (int *)malloc(MROWS*sizeof(int));
  h_costdata = (int *)malloc(MROWS*NCOLS*sizeof(int));
  for (int i=0; i<(MROWS*NCOLS); i++)
    h_costdata[i] = rand()%4;
  cudaMalloc((void **)&d_results, MROWS*sizeof(int));
  cudaMalloc((void **)&d_costdata, MROWS*NCOLS*sizeof(int));
  //copy cost data from host to device
  cudaMemcpy(d_costdata, h_costdata, MROWS*NCOLS*sizeof(int), cudaMemcpyHostToDevice);
  mykernel<<<(MROWS + nTPB - 1)/nTPB, nTPB>>>(d_costdata, MROWS, NCOLS, d_results);
  // copy results back from device to host
  cudaMemcpy(h_results, d_results, MROWS*sizeof(int), cudaMemcpyDeviceToHost);
  for (int i=0; i<MROWS; i++){
    int loc_cost = 0;
    for (int j=0; j<NCOLS; j++) loc_cost += h_costdata[(i*NCOLS)+j];
    printf("cost[%d]: host= %d, device = %d\n", i, loc_cost, h_results[i]);
    }
  }

这假设每一行的成本只是每行中元素的总和。如果您具有其他成本功能，则可以相应地在内核for循环中修改活动。这也假定使用C风格的行主数据存储（1 2 3 4 4 1 2 3 3 4 1 2等）。

This assumes "cost" of each row is just the sum of the elements in each row. If you have a different "cost" function, you can modify the activity in the kernel for-loop accordingly. This also assumes C-style row-major data storage (1 2 3 4 4 1 2 3 3 4 1 2 etc.)

如果您改为使用列主存储（1 4 3等），由于可以完全合并数据读取，因此可以略微提高性能。然后您的内核代码应如下所示：

If you instead use column-major storage (1 4 3 etc.), you can slightly improve the performance, since the data reads can be fully coalesced. Then your kernel code could look like this:

for (int i = 0; i < cols; i++)
  mycost += costdata[(i*rows)+tidx];

您还应该使用正确进行cuda错误检查的规范方法是什么。

编辑：如下面的评论所述，对于行较大的存储情况，在某些情况下，可以通过选择加载来提高内存效率16字节数量而不是基本类型。以下是对任意尺寸和（或多或少）任意基本类型实现此想法的修改版本：

EDIT: As discussed in the comments below, for the row-major storage case, in some situations it might give an increase in memory efficiency by electing to load 16-byte quantities rather than the base type. Following is a modified version that implements this idea for arbitrary dimensions and (more or less) arbitrary base types:

#include <iostream>
#include <typeinfo>
#include <cstdlib>
#include <vector_types.h>

#define MROWS 1742
#define NCOLS 801
#define nTPB 256

typedef double mytype;

__host__ int sizetype(){
  int size = 0;
  if ((typeid(mytype) == typeid(float)) || (typeid(mytype) == typeid(int)) || (typeid(mytype) == typeid(unsigned int)))
      size = 4;
  else if (typeid(mytype) == typeid(double))
      size = 8;
  else if ((typeid(mytype) == typeid(unsigned char)) || (typeid(mytype) == typeid(char)))
      size = 1;
  return size;
  }


template<typename T>
__global__ void mykernel(const T *costdata, int rows, int cols, T *results, int size, size_t pitch){
  int chunk = 16/size;  // assumes size is a factor of 16
  int tidx = threadIdx.x + blockDim.x*blockIdx.x;
  if (tidx < rows){
    T *myrowptr = (T *)(((unsigned char *)costdata) + tidx*pitch);
    T mycost = (T)0;
    int count = 0;
    while (count < cols){
      if ((cols-count)>=chunk){
      // read 16 bytes
        int4 temp = *((int4 *)(myrowptr + count));
        int bcount = 16;
        int j = 0;
        while (bcount > 0){
          mycost += *(((T *)(&temp)) + j++);
          bcount -= size;
          count++;}
        }
      else {
      // read one quantity at a time
        for (; count < cols; count++)
          mycost += myrowptr[count];
        }
    results[tidx] = mycost;
    }
  }
}

int main(){
  int typesize = sizetype();
  if (typesize == 0) {std::cout << "invalid type selected" << std::endl; return 1;}
  //define and initialize host and device storage for cost and results
  mytype *d_costdata, *h_costdata, *d_results, *h_results;
  h_results = (mytype *)malloc(MROWS*sizeof(mytype));
  h_costdata = (mytype *)malloc(MROWS*NCOLS*sizeof(mytype));
  for (int i=0; i<(MROWS*NCOLS); i++)
    h_costdata[i] = (mytype)(rand()%4);
  size_t pitch = 0;
  cudaMalloc((void **)&d_results, MROWS*sizeof(mytype));
  cudaMallocPitch((void **)&d_costdata, &pitch, NCOLS*sizeof(mytype), MROWS);
  //copy cost data from host to device
  cudaMemcpy2D(d_costdata, pitch, h_costdata, NCOLS*sizeof(mytype), NCOLS*sizeof(mytype),  MROWS, cudaMemcpyHostToDevice);

  mykernel<<<(MROWS + nTPB - 1)/nTPB, nTPB>>>(d_costdata, MROWS, NCOLS, d_results, typesize, pitch);
  // copy results back from device to host
  cudaMemcpy(h_results, d_results, MROWS*sizeof(mytype), cudaMemcpyDeviceToHost);
  for (int i=0; i<MROWS; i++){
    mytype loc_cost = (mytype)0;
    for (int j=0; j<NCOLS; j++) loc_cost += h_costdata[(i*NCOLS)+j];
    if ((i < 10) && (typesize > 1))
      std::cout <<"cost[" << i << "]: host= " << loc_cost << ", device = " << h_results[i] << std::endl;
    if (loc_cost != h_results[i]){ std::cout << "mismatch at index" << i << "should be:" << loc_cost << "was:" << h_results[i] << std::endl; return 1; }
    }
  std::cout << "Results are correct!" << std::endl;
  }

这篇关于在CUDA中汇总矩阵的行（以行优先或列优先顺序存储）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！