使用NVIDIA的nvcc编译器编译并生成.cl文件? [英] Compile and build .cl file using NVIDIA's nvcc Compiler?

问题描述

是否可以使用NVIDIA的nvcc编译器来编译.cl文件?我正在尝试设置Visual Studio 2010以在CUDA平台下编码Opencl.但是，当我选择CUDA C/C ++编译器进行编译和生成.cl文件时，它给了我类似nvcc不存在的错误.有什么问题吗?

Is it possible to compile .cl file using NVIDIA's nvcc compiler?? I am trying to set up visual studio 2010 to code Opencl under CUDA platform. But when I select CUDA C/C++ Compiler to compile and build .cl file, it gives me errors like nvcc does not exist. What is the issue?

推荐答案

您应该能够使用nvcc来编译OpenCL代码.通常，我建议对C兼容代码使用文件名扩展名.c，对C ++兼容代码使用文件名扩展名(*)，但是nvcc具有文件名扩展名覆盖选项(-x ...)，这样我们可以修改行为.这是一个使用CUDA 8.0.61，RHEL 7，Tesla K20x的工作示例:

You should be able to use nvcc to compile OpenCL codes. Normally, I would suggest using a filename extension of .c for a C-compliant code, and .cpp for a C++ compliant code(*), however nvcc has filename extension override options (-x ...) so that we can modify the behavior. Here is a worked example using CUDA 8.0.61, RHEL 7, Tesla K20x:

$ 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有意义或有意义的事情，因此，我希望避免出现问题.它仅用于演示符合C ++的OpenCL代码的编译.

Note that the code here doesn't do anything sensible or significant, so I'd prefer to avoid questions. It's just for demonstration of compilation of a C++ compliant OpenCL code.

(*)(因为此类文件也可以由普通的主机编译器(例如gnu编译器)进行处理，并为include和link选项使用适当的开关.)

(*)(Because such files could also be readily processed by an ordinary host compiler, e.g. gnu compilers, with appropriate switches for include and link options.)

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