Question

想象一下带有关联属性的二进制操作（让它命名为“+”）。当您可以并行计算a1 + a2 + a3 + a4 + ...时，首先计算

b1 = a1 + a2
b2 = a3 + a4

然后

c1 = b1 + b2
c2 = b3 + b4

然后对上一步的结果做同样的事情，依此类推，直到剩下一个元素。

我正在学习OpenCL并试图实现这种方法来总结数组中的所有元素。我是这项技术的新手，所以程序可能看起来很奇怪。

这是内核：

__kernel void reduce (__global float *input, __global float *output)
{
    size_t gl = get_global_id (0);
    size_t s = get_local_size (0);
    int i;
    float accum = 0;

    for (i=0; i<s; i++) {
        accum += input[s*gl+i];
    }

    output[gl] = accum;
}

这是主程序：

#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <CL/cl.h>

#define N (64*64*64*64)

#include <sys/time.h>
#include <stdlib.h>

double gettime ()
{
    struct timeval tv;
    gettimeofday (&tv, NULL);
    return (double)tv.tv_sec + (0.000001 * (double)tv.tv_usec);
}

int main()
{
    int i, fd, res = 0;
    void* kernel_source = MAP_FAILED;

    cl_context context;
    cl_context_properties properties[3];
    cl_kernel kernel;
    cl_command_queue command_queue;
    cl_program program;
    cl_int err;
    cl_uint num_of_platforms=0;
    cl_platform_id platform_id;
    cl_device_id device_id;
    cl_uint num_of_devices=0;
    cl_mem input, output;
    size_t global, local;

    cl_float *array = malloc (sizeof (cl_float)*N);
    cl_float *array2 = malloc (sizeof (cl_float)*N);
    for (i=0; i<N; i++) array[i] = i;

    fd = open ("kernel.cl", O_RDONLY);
    if (fd == -1) {
        perror ("Cannot open kernel");
        res = 1;
        goto cleanup;
    }
    struct stat s;

    res = fstat (fd, &s);
    if (res == -1) {
        perror ("Cannot stat() kernel");
        res = 1;
        goto cleanup;
    }

    kernel_source = mmap (NULL, s.st_size, PROT_READ, MAP_PRIVATE, fd, 0);
    if (kernel_source == MAP_FAILED) {
        perror ("Cannot map() kernel");
        res = 1;
        goto cleanup;
    }

    if (clGetPlatformIDs (1, &platform_id, &num_of_platforms) != CL_SUCCESS) {
        printf("Unable to get platform_id\n");
        res = 1;
        goto cleanup;
    }

    if (clGetDeviceIDs(platform_id, CL_DEVICE_TYPE_GPU, 1, &device_id,
                       &num_of_devices) != CL_SUCCESS)
    { 
        printf("Unable to get device_id\n");
        res = 1;
        goto cleanup;
    }
    properties[0]= CL_CONTEXT_PLATFORM;
    properties[1]= (cl_context_properties) platform_id;
    properties[2]= 0;
    context = clCreateContext(properties,1,&device_id,NULL,NULL,&err);
    command_queue = clCreateCommandQueue(context, device_id, 0, &err);
    program = clCreateProgramWithSource(context, 1, (const char**)&kernel_source, NULL, &err);


    if (clBuildProgram(program, 0, NULL, NULL, NULL, NULL) != CL_SUCCESS) {
        char buffer[4096];
        size_t len;

        printf("Error building program\n");
        clGetProgramBuildInfo (program, device_id, CL_PROGRAM_BUILD_LOG, sizeof (buffer), buffer, &len);
        printf ("%s\n", buffer);
        res = 1;
        goto cleanup;
     }

    kernel = clCreateKernel(program, "reduce", &err);
    if (err != CL_SUCCESS) {
        printf("Unable to create kernel\n");
        res = 1;
        goto cleanup;
    }

    // create buffers for the input and ouput
    input = clCreateBuffer(context, CL_MEM_READ_ONLY, 
                            sizeof(cl_float) * N, NULL, NULL);
    output = clCreateBuffer(context, CL_MEM_WRITE_ONLY, 
                            sizeof(cl_float) * N, NULL, NULL);

    // load data into the input buffer
    clEnqueueWriteBuffer(command_queue, input, CL_TRUE, 0, 
                          sizeof(cl_float) * N, array, 0, NULL, NULL);

    size_t size = N;
    cl_mem tmp;
    double time = gettime();
    while (size > 1)
    {
        // set the argument list for the kernel command
        clSetKernelArg(kernel, 0, sizeof(cl_mem), &input);
        clSetKernelArg(kernel, 1, sizeof(cl_mem), &output);
        global = size;
        local = 64;

        // enqueue the kernel command for execution
        clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global, 
                           &local, 0, NULL, NULL);
        clFinish(command_queue);
        size = size/64;
        tmp = output;
        output = input;
        input = tmp;
    }
    cl_float answer[1];
    clEnqueueReadBuffer(command_queue, tmp, CL_TRUE, 0, 
                        sizeof(cl_float), array, 0, NULL, NULL);
    time = gettime() - time;
    printf ("%f %f\n", array[0], time);

cleanup:
    free (array);
    free (array2);
    clReleaseMemObject(input);
    clReleaseMemObject(output);
    clReleaseProgram(program);
    clReleaseKernel(kernel);
    clReleaseCommandQueue(command_queue);
    clReleaseContext(context);

    if (kernel_source != MAP_FAILED) munmap (kernel_source, s.st_size);
    if (fd != -1) close (fd);

    _Exit (res); // Kludge
    return res;
}

所以我重新运行内核，直到缓冲区中只有一个元素。这是计算OpenCL中元素总和的正确方法吗？在CPU（编译的clang 4.0.0和gettime标志）上执行一个简单循环的时间时，用-O2 -ffast-math测量的时间大约慢10倍。我使用的硬件：Amd Ryzen 5 1600X和Amd Radeon HD 6950。

Answer 1

您可以采取一些措施来提高效果。

首先，摆脱循环内的/usr/local/调用。这迫使内核的单独执行依赖于命令队列的整个状态，在继续之前到达与主机的同步点，这是不必要的。唯一需要的同步是内核按顺序执行，即使你有一个无序队列（你的程序也没有请求它），你可以保证简单地使用事件对象。

clFinish

可能要考虑的另一件事是在一个内核中执行减少，而不是在同一内核的多个调用中进行扩展。 This is one potential例子，虽然它可能比你需要的更复杂。

在OpenCL中减少最佳做法是什么？

1 个答案: