在OpenCL中減少的最佳做法是什麼？

Question

想象一下使用關聯屬性的二元運算（讓它命名爲「+」）。當你可以計算並行a1 + a2 + a3 + a4 + ...，第一計算

b1 = a1 + a2 
b2 = a3 + a4 

然後

c1 = b1 + b2 
c2 = b3 + b4 

然後做同樣的事情上一步的結果，依此類推，直到還剩下一個元素。

我在學習OpenCL並嘗試實現這種方法來總結數組中的所有元素。我是這個技術的全新手，所以這個程序可能看起來很奇怪。

這是內核：

__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; 
} 

這是主程序：

#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; 
} 

所以我重新運行的內核，直到有隻有一個緩衝件。這是計算OpenCL中元素總和的正確方法嗎？我用gettime測量的時間比CPU上一個簡單循環的執行時間（編譯鐺4.0.0和-O2 -ffast-math標誌）慢大約10倍。我使用的硬件：Amd Ryzen 5 1600X和Amd Radeon HD 6950.

Answer 1

有幾件事你可以試着改善性能。

首先，擺脫您的循環內的clFinish調用。這會迫使內核的單獨執行取決於命令隊列與主機達到同步點的整個狀態，然後才能繼續，這是不必要的。唯一需要的同步是內核按順序執行，即使你有一個亂序隊列（你的程序沒有請求），你可以保證簡單地使用事件對象。

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); 

另一件可能研究的內容是在一個內核中執行縮減操作，而不是在同一個內核的多個調用中執行縮減操作。 This is one potential示例，但它可能比您需要的更復雜。