使用 NVIDIA 的 nvcc 编译器编译和构建 .cl 文件？

Question

是否可以使用 NVIDIA 的 nvcc 编译器编译 .cl 文件？我正在尝试设置 Visual Studio 2010 在 CUDA 平台下编写 Opencl 代码。但是当我选择 CUDA C/C++ Compiler 来编译和构建 .cl 文件时，它给了我类似 nvcc 不存在的错误。有什么问题？

Answer 1

您应该可以使用nvcc 编译 OpenCL 代码。通常，我建议使用 .c 的文件扩展名作为 C 兼容代码，.cpp 用于 C++ 兼容代码 (*)，但是 nvcc 具有文件扩展名覆盖选项 (-x ...)，以便我们可以修改行为。这是一个使用 CUDA 8.0.61、RHEL 7、Tesla K20x 的工作示例：

$ cat t4.cpp
#include <CL/opencl.h>
#include <stdint.h>
#include <stdio.h>
#include <inttypes.h>
#include <stdlib.h>

const char source[] =
"__kernel void test_rotate(__global ulong *d_count, ulong loops, ulong patt)"
"{"
"  ulong n = patt;"
"  for (ulong i = 0; i<loops; i++)"
"    n &= (107 << (patt+(i%7)));"
"  d_count[0] = n + loops;"
"}"
;

int main(int argc, char *argv[])
{
  cl_platform_id platform;
  cl_device_id device;
  cl_context context;
  cl_command_queue queue1, queue2;
  cl_program program;
  cl_mem mem1, mem2;
  cl_kernel kernel;

  bool two_kernels = false;
  unsigned long long loops = 1000;
  if (argc > 1) loops *= atoi(argv[1]);
  if (argc > 2) two_kernels = true;
  if (two_kernels) printf("running two kernels\n");
  else printf("running one kernel\n");
  printf("running  %lu loops\n", loops);
  unsigned long long pattern = 1;
  clGetPlatformIDs(1, &platform, NULL);
  clGetDeviceIDs(platform, CL_DEVICE_TYPE_ALL, 1, &device, NULL);
  context = clCreateContext(NULL, 1, &device, NULL, NULL, NULL);
  queue1 = clCreateCommandQueue(context, device, CL_QUEUE_OUT_OF_ORDER_EXEC_MODE_ENABLE, NULL);
  queue2 = clCreateCommandQueue(context, device, CL_QUEUE_OUT_OF_ORDER_EXEC_MODE_ENABLE, NULL);

  const char *sources[1] = {source};
  program = clCreateProgramWithSource(context, 1, sources, NULL, NULL);
  clBuildProgram(program, 1, &device, NULL, NULL, NULL);
  mem1 = clCreateBuffer(context, CL_MEM_READ_WRITE, 1*sizeof(cl_ulong), NULL, NULL);
  mem2 = clCreateBuffer(context, CL_MEM_READ_WRITE, 1*sizeof(cl_ulong), NULL, NULL);
  kernel = clCreateKernel(program, "test_rotate", NULL);
  const size_t work_size[1] = {1};
  clSetKernelArg(kernel, 0, sizeof(mem1), &mem1);
  clSetKernelArg(kernel, 1, sizeof(loops), &loops);
  clSetKernelArg(kernel, 2, sizeof(pattern), &pattern);

  clEnqueueNDRangeKernel(queue1, kernel, 1, NULL, work_size, work_size, 0, NULL, NULL);
  if (two_kernels){
    clSetKernelArg(kernel, 0, sizeof(mem2), &mem2);
    clSetKernelArg(kernel, 1, sizeof(loops), &loops);
    clSetKernelArg(kernel, 2, sizeof(pattern), &pattern);

    clEnqueueNDRangeKernel(queue2, kernel, 1, NULL, work_size, work_size, 0, NULL, NULL);
    }
  cl_ulong *buf1 = (cl_ulong *)clEnqueueMapBuffer(queue1, mem1, true, CL_MAP_READ, 0, 1*sizeof(cl_ulong), 0, NULL, NULL, NULL);
  cl_ulong *buf2 = (cl_ulong *)clEnqueueMapBuffer(queue2, mem2, true, CL_MAP_READ, 0, 1*sizeof(cl_ulong), 0, NULL, NULL, NULL);
  printf("result1: %lu\n", buf1[0]);
  printf("result2: %lu\n", buf2[0]);
  clEnqueueUnmapMemObject(queue1, mem1, buf1, 0, NULL, NULL);
  clEnqueueUnmapMemObject(queue2, mem2, buf2, 0, NULL, NULL);
  return 0;
}
$ nvcc -arch=sm_35 -o t4 t4.cpp -lOpenCL
$ ./t4
running one kernel
running  1000 loops
result1: 1000
result2: 0
$ cp t4.cpp t4.cl
$ nvcc -arch=sm_35 -x cu -o t4 t4.cl -lOpenCL
$ ./t4
running one kernel
running  1000 loops
result1: 1000
result2: 0
$

请注意，这里的代码没有做任何明智或重要的事情，所以我宁愿避免提问。这只是为了演示编译符合 C++ 标准的 OpenCL 代码。

(*)（因为此类文件也可以由普通主机编译器轻松处理，例如 gnu 编译器，带有适当的包含和链接选项开关。）