GPGPU OpenCL Reduction操作与group同步

Reduction操作：规约操作就是由多个数生成一个数，如求最大值、最小值、向量点积、求和等操作，都属于这一类操作。

有大量数据的情况下，使用GPU进行任务并行与数据并行，可以收到可好的效果。

group同步：OpenCL只提供了工作组内的各线程之间的同步机制，并没有提供所有线程的同步。提供组内item-work同步的方法：

　　void barrier (cl_mem_fence_flags flags) 

　　参数说明：cl_mem_fence_flags 可以取CLK_LOCAL_MEM_FENCE、CLK_GLOBAL_MEM_FENCE

　　函数说明：(1)一个work-group中所有work-item遇到barrier方法，都要等待其他work-item也到达该语句，才能执行后面的程序；

　　　　　　　 (2)还可以组内的work-item对local or global memory的顺序读写操作。

 

如下图中每个大框表示任务并行、每个group线程；框中的计算是数据并行、每个item-work线程：

作为练习，给出个完整的使用OpenCL计算整数序列求和，在数据并行中使用Local Memory 加速，group组内并行同步使用CLK_LOCAL_MEM_FENCE。

程序实例(整数序列求和)：

1.核函数(Own_Reduction_Kernels.cl)：

 1 __kernel

 2 void 

 3 reduce(__global uint4* input, __global uint4* output, int NUM)

 4 {    

 5     NUM = NUM / 4;    //每四个数为一个整体uint4。

 6     unsigned int tid = get_local_id(0);

 7     unsigned int localSize = get_local_size(0);

 8     unsigned int globalSize = get_global_size(0);

 9 

10     uint4 res=(uint4){0,0,0,0};

11     __local uint4 resArray[64];

12 

13     

14     unsigned int i = get_global_id(0);

15     while(i < NUM)

16     {

17         res+=input[i];

18         i+=globalSize;

19     }

20     resArray[tid]=res;    //将每个work-item计算结果保存到对应__local memory中

21     barrier(CLK_LOCAL_MEM_FENCE);

22 

23     // do reduction in shared mem

24     for(unsigned int s = localSize >> 1; s > 0; s >>= 1) 

25     {

26         if(tid < s) 

27         {

28             resArray[tid] += resArray[tid + s];

29         }

30         barrier(CLK_LOCAL_MEM_FENCE);

31     }

32 

33     // write result for this block to global mem

34     if(tid == 0) 

35         output[get_group_id(0)] = resArray[0];

36 }

2.tool.h 、tool.cpp

 见：http://www.cnblogs.com/xudong-bupt/p/3582780.html 

3.Reduction.cpp

  1 #include <CL/cl.h>

  2 #include "tool.h"

  3 #include <string.h>

  4 #include <stdio.h>

  5 #include <stdlib.h>

  6 #include <iostream>

  7 #include <string>

  8 #include <fstream>

  9 using namespace std;

 10 

 11 int isVerify(int NUM,int groupNUM,int *res)    //校验结果

 12 {

 13        int sum1 = (NUM+1)*NUM/2;

 14     int sum2 = 0;

 15     for(int i = 0;i < groupNUM*4; i++)

 16         sum2 += res[i];

 17     if(sum1 == sum2)

 18         return 0;

 19     return -1;

 20 }

 21 

 22 void isStatusOK(cl_int status)    //判断状态码

 23 {

 24     if(status == CL_SUCCESS)

 25         cout<<"RIGHT"<<endl;

 26     else

 27         cout<<"ERROR"<<endl;

 28 }

 29 

 30 int main(int argc, char* argv[])

 31 {

 32     cl_int    status;

 33     /**Step 1: Getting platforms and choose an available one(first).*/

 34     cl_platform_id platform;

 35     getPlatform(platform);

 36 

 37     /**Step 2:Query the platform and choose the first GPU device if has one.*/

 38     cl_device_id *devices=getCl_device_id(platform);

 39 

 40     /**Step 3: Create context.*/

 41     cl_context context = clCreateContext(NULL,1, devices,NULL,NULL,NULL);

 42 

 43     /**Step 4: Creating command queue associate with the context.*/

 44     cl_command_queue commandQueue = clCreateCommandQueue(context, devices[0], 0, NULL);

 45 

 46     /**Step 5: Create program object */

 47     const char *filename = "Own_Reduction_Kernels.cl";

 48     string sourceStr;

 49     status = convertToString(filename, sourceStr);

 50     const char *source = sourceStr.c_str();

 51     size_t sourceSize[] = {strlen(source)};

 52     cl_program program = clCreateProgramWithSource(context, 1, &source, sourceSize, NULL);

 53 

 54     /**Step 6: Build program. */

 55     status=clBuildProgram(program, 1,devices,NULL,NULL,NULL);

 56 

 57     /**Step 7: Initial input,output for the host and create memory objects for the kernel*/

 58     int NUM=25600;    //6400*4

 59     size_t global_work_size[1] = {640};  ///

 60     size_t local_work_size[1]={64};    ///256 PE

 61     size_t groupNUM=global_work_size[0]/local_work_size[0];

 62     int* input = new int[NUM];

 63     for(int i=0;i<NUM;i++)

 64         input[i]=i+1;

 65     int* output = new int[(global_work_size[0]/local_work_size[0])*4];

 66 

 67     cl_mem inputBuffer = clCreateBuffer(context, CL_MEM_READ_ONLY|CL_MEM_COPY_HOST_PTR, (NUM) * sizeof(int),(void *) input, NULL);

 68     cl_mem outputBuffer = clCreateBuffer(context, CL_MEM_WRITE_ONLY , groupNUM*4* sizeof(int), NULL, NULL);

 69 

 70     /**Step 8: Create kernel object */

 71     cl_kernel kernel = clCreateKernel(program,"reduce", NULL);

 72 

 73     /**Step 9: Sets Kernel arguments.*/

 74     status = clSetKernelArg(kernel, 0, sizeof(cl_mem), (void *)&inputBuffer);

 75     status = clSetKernelArg(kernel, 1, sizeof(cl_mem), (void *)&outputBuffer);

 76     status = clSetKernelArg(kernel, 2, sizeof(int), &NUM);

 77 

 78     /**Step 10: Running the kernel.*/

 79     cl_event enentPoint;

 80     status = clEnqueueNDRangeKernel(commandQueue, kernel, 1, NULL, global_work_size, local_work_size, 0, NULL, &enentPoint);

 81     clWaitForEvents(1,&enentPoint); ///wait

 82     clReleaseEvent(enentPoint);

 83     isStatusOK(status);

 84             

 85     /**Step 11: Read the cout put back to host memory.*/

 86     status = clEnqueueReadBuffer(commandQueue, outputBuffer, CL_TRUE, 0,groupNUM*4 * sizeof(int), output, 0, NULL, NULL);

 87     isStatusOK(status);

 88     if(isVerify(NUM, groupNUM ,output) == 0)

 89         cout<<"The result is right!!!"<<endl;

 90     else

 91         cout<<"The result is wrong!!!"<<endl;

 92 

 93     /**Step 12: Clean the resources.*/

 94     status = clReleaseKernel(kernel);//*Release kernel.

 95     status = clReleaseProgram(program);    //Release the program object.

 96     status = clReleaseMemObject(inputBuffer);//Release mem object.

 97     status = clReleaseMemObject(outputBuffer);

 98     status = clReleaseCommandQueue(commandQueue);//Release  Command queue.

 99     status = clReleaseContext(context);//Release context.

100 

101     free(input);

102     free(output);

103     free(devices);

104     return 0;

105 }

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