使用JOCL / OPENCL计算强度的加速和

时间:2012-11-24 16:46:53

标签: java opencl jocl

您好我是JOCL(opencl)的新手。我编写了这段代码来获取每张图像强度的总和。内核采用所有图像的所有像素的一维数组。图像为300x300,因此每张图像为90000像素。目前它比我按顺序执行的速度慢。

我的代码

package PAR;

/*
 * JOCL - Java bindings for OpenCL
 * 
 * Copyright 2009 Marco Hutter - http://www.jocl.org/
 */
import IMAGE_IO.ImageReader;
import IMAGE_IO.Input_Folder;
import static org.jocl.CL.*;

import org.jocl.*;

/**
 * A small JOCL sample.
 */
public class IPPARA {

    /**
     * The source code of the OpenCL program to execute
     */
    private static String programSource =
            "__kernel void "
            + "sampleKernel(__global uint *a,"
            + "             __global uint *c)"
            + "{"
            + "__private uint intensity_core=0;"
            + "      uint i = get_global_id(0);"
            + "      for(uint j=i*90000; j < (i+1)*90000; j++){ "
            + "              intensity_core += a[j];"
            + "     }"
            + "c[i]=intensity_core;" 
            + "}";

    /**
     * The entry point of this sample
     *
     * @param args Not used
     */
    public static void main(String args[]) {
        long numBytes[] = new long[1];

        ImageReader imagereader = new ImageReader() ;
        int srcArrayA[]  = imagereader.readImages();

        int size[] = new int[1];
        size[0] = srcArrayA.length;
        long before = System.nanoTime();
        int dstArray[] = new int[size[0]/90000];


        Pointer srcA = Pointer.to(srcArrayA);
        Pointer dst = Pointer.to(dstArray);


        // Obtain the platform IDs and initialize the context properties
        System.out.println("Obtaining platform...");
        cl_platform_id platforms[] = new cl_platform_id[1];
        clGetPlatformIDs(platforms.length, platforms, null);
        cl_context_properties contextProperties = new cl_context_properties();
        contextProperties.addProperty(CL_CONTEXT_PLATFORM, platforms[0]);

        // Create an OpenCL context on a GPU device
        cl_context context = clCreateContextFromType(
                contextProperties, CL_DEVICE_TYPE_CPU, null, null, null);
        if (context == null) {
            // If no context for a GPU device could be created,
            // try to create one for a CPU device.
            context = clCreateContextFromType(
                    contextProperties, CL_DEVICE_TYPE_CPU, null, null, null);

            if (context == null) {
                System.out.println("Unable to create a context");
                return;
            }
        }

        // Enable exceptions and subsequently omit error checks in this sample
        CL.setExceptionsEnabled(true);

        // Get the list of GPU devices associated with the context
        clGetContextInfo(context, CL_CONTEXT_DEVICES, 0, null, numBytes);

        // Obtain the cl_device_id for the first device
        int numDevices = (int) numBytes[0] / Sizeof.cl_device_id;
        cl_device_id devices[] = new cl_device_id[numDevices];
        clGetContextInfo(context, CL_CONTEXT_DEVICES, numBytes[0],
                Pointer.to(devices), null);

        // Create a command-queue
        cl_command_queue commandQueue =
                clCreateCommandQueue(context, devices[0], 0, null);

        // Allocate the memory objects for the input- and output data
        cl_mem memObjects[] = new cl_mem[2];
        memObjects[0] = clCreateBuffer(context,
                CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
                Sizeof.cl_uint * srcArrayA.length, srcA, null);
        memObjects[1] = clCreateBuffer(context,
                CL_MEM_READ_WRITE,
                Sizeof.cl_uint * (srcArrayA.length/90000), null, null);

        // Create the program from the source code
        cl_program program = clCreateProgramWithSource(context,
                1, new String[]{programSource}, null, null);

        // Build the program
        clBuildProgram(program, 0, null, null, null, null);

        // Create the kernel
        cl_kernel kernel = clCreateKernel(program, "sampleKernel", null);

        // Set the arguments for the kernel
        clSetKernelArg(kernel, 0,
                Sizeof.cl_mem, Pointer.to(memObjects[0]));
        clSetKernelArg(kernel, 1,
                Sizeof.cl_mem, Pointer.to(memObjects[1]));

        // Set the work-item dimensions
        long local_work_size[] = new long[]{1};
        long global_work_size[] = new long[]{(srcArrayA.length/90000)*local_work_size[0]};


        // Execute the kernel
        clEnqueueNDRangeKernel(commandQueue, kernel, 1, null,
                global_work_size, local_work_size, 0, null, null);

        // Read the output data
        clEnqueueReadBuffer(commandQueue, memObjects[1], CL_TRUE, 0,
                (srcArrayA.length/90000) * Sizeof.cl_float, dst, 0, null, null);

        // Release kernel, program, and memory objects
        clReleaseMemObject(memObjects[0]);
        clReleaseMemObject(memObjects[1]);
        clReleaseKernel(kernel);
        clReleaseProgram(program);
        clReleaseCommandQueue(commandQueue);
        clReleaseContext(context);


        long after = System.nanoTime();

        System.out.println("Time: " + (after - before) / 1e9);

    }
}

在答案中的建议之后,通过CPU的并行代码几乎与顺序代码一样快。是否还有其他改进措施?

2 个答案:

答案 0 :(得分:2)

 for(uint j=i*90000; j < (i+1)*90000; j++){ "
        + "              c[i] += a[j];"

1)你正在使用全局内存(c [])求和,这很慢。使用私有变量使其更快。  像这样:

          "__kernel void "
        + "sampleKernel(__global uint *a,"
        + "             __global uint *c)"
        + "{"
        + "__private uint intensity_core=0;" <---this is a private variable of each core
        + "      uint i = get_global_id(0);"
        + "      for(uint j=i*90000; j < (i+1)*90000; j++){ "
        + "              intensity_core += a[j];" <---register is at least 100x faster than global memory
         //but we cannot get rid of a[] so the calculation time cannot be less than %50
        + "     }"
        + "c[i]=intensity_core;"   
        + "}";  //expecting %100 speedup

现在你有c [图像数]数组的强度和。

您的本地工作规模为1,如果您有至少160张图片(这是您的gpu的核心编号),那么计算将使用所有核心。

您需要90000 * num_images次读取和num_images写入以及90000 * num_images寄存器读/写。使用寄存器会使内核时间缩短一半。

2)每2个内存访问只进行1次数学运算。每1个内存访问至少需要10个数学,才能使用gpu的一小部分峰值Gflops(6490M时为250 Gflops峰值)

你的i7 cpu可以轻松拥有100 Gflops,但你的记忆力将成为瓶颈。通过pci-express发送整个数据时情况更糟(HD Graphics 3000的额定值为125 GFLOPS)

 // Obtain a device ID 
    cl_device_id devices[] = new cl_device_id[numDevices];
    clGetDeviceIDs(platform, deviceType, numDevices, devices, null);
    cl_device_id device = devices[deviceIndex];
 //one of devices[] element must be your HD3000.Example: devices[0]->gpu devices[1]->cpu 
 //devices[2]-->HD3000

在你的计划中:

 // Obtain the cl_device_id for the first device
    int numDevices = (int) numBytes[0] / Sizeof.cl_device_id;
    cl_device_id devices[] = new cl_device_id[numDevices];
    clGetContextInfo(context, CL_CONTEXT_DEVICES, numBytes[0],
            Pointer.to(devices), null);

将第一个设备概率地称为gpu。

答案 1 :(得分:0)

您应该按照300x300图像使用整个工作组。这将有助于使gpu内核饱和并让您使用本地内存。内核还应该能够像设备上的计算单元一样同时处理任意数量的图像。

下面的内核将您的减少分为三个步骤。

  1. 将每个工作项的值读入一个私人单位
  2. 将私有var写入本地内存(非常简单,但很重要)
  3. 减少本地内存中的值以获取最终值。这里有两种方法可以做到这一点。
  4. 定义了WG_MAX_SIZE,因为我不是传递可变大小的本地内存块的粉丝。该值为64,因为这是在大多数平台上使用的良好值。如果要试验较大的工作组,请确保将此值设置得更高。小于WG_MAX_SIZE的工作组仍然可以正常工作。

    #define WORK_SIZE 90000
    #define WG_MAX_SIZE 64
    __kernel void sampleKernel(__global uint *a, __global uint *c)
    {
    
        local uint intensity_core[WG_MAX_SIZE];
        private uint workItemIntensity = 0;
    
        int gid = get_group_id(0);
        int lid = get_local_id(0);
        int wgsize = get_local_size(0);
        int i;
    
        for(i=gid*WORK_SIZE; i < (gid+1)*WORK_SIZE; i+=wgsize){ 
            workItemIntensity += a[j];
        }
        intensity_core[lid] = workItemIntensity;
        mem_fence(CLK_LOCAL_MEM_FENCE);
    
        //option #1
        //loop to reduce the final values O(n) time
        if(lid == 0){
            for(i=1;i<wgsize;i++){
                workItemIntensity += intensity_core[i];
            }
            c[gid]=intensity_core;
        }
    
        //option #2
        //O(logn) time reduction
        //assumes work group size is a power of 2
        int steps = 32 - clz(wgsize);
        for(i=1;i<steps;i++){
            if(lid % (1 << i) == 0){
                intensity_core[lid] += intensity_core[i<<(i-1)];
            }
            mem_fence(CLK_LOCAL_MEM_FENCE);
        }
        if(lid == 0){
            c[gid]=intensity_core[0];
        }
    }