java - 使用 JOCL/OPENCL 加速强度总和计算

Question

您好，我是 JOCL (opencl) 的新手。我编写这段代码是为了获取每张图像的强度总和。内核采用一个一维数组，其中包含所有图像的所有像素，这些像素彼此放在一起。一张图片是 300x300 ，所以每张图片有 90000 像素。目前它比我按顺序执行此操作时慢。

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);

    }
}

根据答案中的建议，通过 CPU 的并行代码几乎与顺序代码一样快。是否还有更多可以改进的地方？

Answer 1

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

1) 您正在使用全局内存 (c[]) 求和，这很慢。使用私有(private)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[图像数量] 个强度总和数组。

你的 local-work-size 是 1 那么如果你有至少 160 张图像(这是你的 gpu 的核心数)那么计算将使用所有核心。

您将需要 90000*num_images 次读取和 num_images 次写入以及 90000*num_images 寄存器读/写。使用寄存器将使您的内核时间减半。

2) 你每 2 次内存访问只做 1 次数学运算。每次内存访问至少需要 10 个数学运算才能使用 gpu 峰值 Gflops 的一小部分(6490M 峰值为 250 Gflops)

您的 i7 cpu 可以轻松达到 100 Gflops，但您的内存将成为瓶颈。当您通过 pci-express 发送整个数据时，情况更糟。(HD Graphics 3000 额定为 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

在你的程序中:

 // 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);

第一个设备可能是 gpu。