在OpenCL中减少的最佳做法是什么？

Question

想象一下使用关联属性的二元运算（让它命名为“+”）。当你可以计算并行a1 + a2 + a3 + a4 + ...，第一计算

b1 = a1 + a2 
b2 = a3 + a4 

然后

c1 = b1 + b2 
c2 = b3 + b4 

然后做同样的事情上一步的结果，依此类推，直到还剩下一个元素。

我在学习OpenCL并尝试实现这种方法来总结数组中的所有元素。我是这个技术的全新手，所以这个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; 
} 

所以我重新运行的内核，直到有只有一个缓冲件。这是计算OpenCL中元素总和的正确方法吗？我用gettime测量的时间比CPU上一个简单循环的执行时间（编译铛4.0.0和-O2 -ffast-math标志）慢大约10倍。我使用的硬件：Amd Ryzen 5 1600X和Amd Radeon HD 6950.

Answer 1

有几件事你可以试着改善性能。

首先，摆脱您的循环内的clFinish调用。这会迫使内核的单独执行取决于命令队列与主机达到同步点的整个状态，然后才能继续，这是不必要的。唯一需要的同步是内核按顺序执行，即使你有一个乱序队列（你的程序没有请求），你可以保证简单地使用事件对象。

size_t size = N; 
size_t total_expected_events = 0; 
for(size_t event_count = size; event_count > 1; event_count /= 64) 
    total_expected_events++; 
cl_event * events = malloc(total_expected_events * sizeof(cl_event)); 
cl_mem tmp; 
double time = gettime(); 
size_t event_index = 0; 
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; 

    if(event_index == 0) 
     // enqueue the kernel command for execution 
     clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global, 
          &local, 0, NULL, events); 
    else 
     clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global, 
          &local, 1, events + (event_index - 1), events + event_index); 
    size = size/64; 
    tmp = output; 
    output = input; 
    input = tmp; 
    event_index++; 
} 
clFinish(command_queue); 
for(; event_index > 0; event_index--) 
    clReleaseEvent(events[event_index-1]); 
free(events); 
cl_float answer[1]; 
clEnqueueReadBuffer(command_queue, tmp, CL_TRUE, 0, 
        sizeof(cl_float), array, 0, NULL, NULL); 

另一件可能研究的内容是在一个内核中执行缩减操作，而不是在同一个内核的多个调用中执行缩减操作。 This is one potential示例，但它可能比您需要的更复杂。