OpenCL入门:(二:用GPU计算两个数组和)

本文编写一个计算两个数组和的程序,用CPU和GPU分别运算,计算运算时间,并且校验最后的运算结果。文中代码偏多,原理建议阅读下面文章,文中介绍了OpenCL相关名词概念。

http://opencl.codeplex.com/wikipage?title=OpenCL%20Tutorials%20-%201  (英文版)

http://www.cnblogs.com/leiben/archive/2012/06/05/2536508.html (博友翻译的中文版)

一、创建工程

按照OpenCL入门:(一:Intel核心显卡OpenCL环境搭建)的创建一个名为OpenCLSum的工程,并且添加一个OpenCLSum.cpp文件,一个OpenCLSum.cl文件(添加时选择添加OpenCL文件)。

二、CPU计算代码

用CPU求两个数组和的代码如下:

void RunAsCpu(
const float *nums1,
const float *nums2,
float* sum,
const int num)
{
for (int i = 0; i < num; i++)
{
sum[i] = nums1[i] + nums2[i];
}
}

三、GPU计算代码

在cl文件中添加如下代码,

//因为运行这个kernel时需要设置一个线程数目,
//所以每个线程都会调用一次这个函数,只需要使
//用get_global_id获取它的线程id就可以求和了
__kernel void RunAsGpu(
__global const float *nums1,
__global const float *nums2,
__global float* sum)
{
int id = get_global_id(0);
sum[id] = nums1[id] + nums2[id];
}

四、主函数流程

流程请参考本文开始推荐的文章,有详细说明,下面只在注释中简单说明

//计时函数
double time_stamp()
{
LARGE_INTEGER curclock;
LARGE_INTEGER freq;
if (
!QueryPerformanceCounter(&curclock) ||
!QueryPerformanceFrequency(&freq)
)
{
return -1;
} return double(curclock.QuadPart) / freq.QuadPart;
}
#define OPENCL_CHECK_ERRORS(ERR) \
if(ERR != CL_SUCCESS) \
{ \
cerr \
<< "OpenCL error with code " << ERR \
<< " happened in file " << __FILE__ \
<< " at line " << __LINE__ \
<< ". Exiting...\n"; \
exit(1); \
}
int main(int argc, const char** argv)
{
cl_int error = 0; // Used to handle error codes
cl_context context;
cl_command_queue queue;
cl_device_id device; // 遍历系统中所有OpenCL平台
cl_uint num_of_platforms = 0;
// 得到平台数目
error = clGetPlatformIDs(0, 0, &num_of_platforms);
OPENCL_CHECK_ERRORS(error);
cout << "可用平台数: " << num_of_platforms << endl; cl_platform_id* platforms = new cl_platform_id[num_of_platforms];
// 得到所有平台的ID
error = clGetPlatformIDs(num_of_platforms, platforms, 0);
OPENCL_CHECK_ERRORS(error);
//遍历平台,选择一个Intel平台的
cl_uint selected_platform_index = num_of_platforms;
for (cl_uint i = 0; i < num_of_platforms; ++i)
{
size_t platform_name_length = 0;
error = clGetPlatformInfo(
platforms[i],
CL_PLATFORM_NAME,
0,
0,
&platform_name_length
);
OPENCL_CHECK_ERRORS(error); // 调用两次,第一次是得到名称的长度
char* platform_name = new char[platform_name_length];
error = clGetPlatformInfo(
platforms[i],
CL_PLATFORM_NAME,
platform_name_length,
platform_name,
0
);
OPENCL_CHECK_ERRORS(error); cout << " [" << i << "] " << platform_name; if (
strstr(platform_name, "Intel") &&
selected_platform_index == num_of_platforms // have not selected yet
)
{
cout << " [Selected]";
selected_platform_index = i;
} cout << endl;
delete[] platform_name;
}
if (selected_platform_index == num_of_platforms)
{
cerr
<< "没有找到Intel平台\n";
return 1;
}
// Device
cl_platform_id platform = platforms[selected_platform_index];
error = clGetDeviceIDs(platform, CL_DEVICE_TYPE_GPU, 1, &device, NULL);
OPENCL_CHECK_ERRORS(error) // Context
context = clCreateContext(0, 1, &device, NULL, NULL, &error);
OPENCL_CHECK_ERRORS(error) // Command-queue
queue = clCreateCommandQueue(context, device, 0, &error);
OPENCL_CHECK_ERRORS(error) //下面初始化测试数据(主机数据)
const int size = 38888888;//大小和内存有关,仅作示例
float* nums1_h = new float[size];
float* nums2_h = new float[size];
float* sum_h = new float[size];
// Initialize both vectors
for (int i = 0; i < size; i++) {
nums1_h[i] = nums2_h[i] = (float)i;
}
//初始化设备数据
const int mem_size = sizeof(float)*size;
// 标志位表示数据只读,并且从nums1_h和nums2_h复制数据
cl_mem nums1_d = clCreateBuffer(context, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR, mem_size, nums1_h, &error);
OPENCL_CHECK_ERRORS(error)
cl_mem nums2_d = clCreateBuffer(context, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR, mem_size, nums2_h, &error);
OPENCL_CHECK_ERRORS(error)
cl_mem sum_d = clCreateBuffer(context, CL_MEM_WRITE_ONLY, mem_size, NULL, &error);
OPENCL_CHECK_ERRORS(error) //读取OpenCLSum.cl文件内容 FILE* fp = fopen("OpenCLSum.cl", "rb");
fseek(fp, 0, SEEK_END);
size_t src_size = ftell(fp);
fseek(fp, 0, SEEK_SET);
const char* source = new char[src_size];
fread((void*)source, 1, src_size, fp);
fclose(fp); //创建编译运行kernel函数
cl_program program = clCreateProgramWithSource(context, 1, &source, &src_size, &error);
OPENCL_CHECK_ERRORS(error)
delete[] source; // Builds the program
error = clBuildProgram(program, 1, &device, NULL, NULL, NULL);
OPENCL_CHECK_ERRORS(error) // Shows the log
char* build_log;
size_t log_size;
// First call to know the proper size
clGetProgramBuildInfo(program, device, CL_PROGRAM_BUILD_LOG, 0, NULL, &log_size);
build_log = new char[log_size + 1];
// Second call to get the log
clGetProgramBuildInfo(program, device, CL_PROGRAM_BUILD_LOG, log_size, build_log, NULL);
build_log[log_size] = '\0';
cout << build_log << endl;
delete[] build_log; // Extracting the kernel
cl_kernel run_as_gpu = clCreateKernel(program, "RunAsGpu", &error);
OPENCL_CHECK_ERRORS(error) //运行kernel程序 // Enqueuing parameters
// Note that we inform the size of the cl_mem object, not the size of the memory pointed by it
error = clSetKernelArg(run_as_gpu, 0, sizeof(cl_mem), &nums1_d);
error |= clSetKernelArg(run_as_gpu, 1, sizeof(cl_mem), &nums2_d);
error |= clSetKernelArg(run_as_gpu, 2, sizeof(cl_mem), &sum_d);
OPENCL_CHECK_ERRORS(error) // Launching kernel
size_t global_work_size = size;
cout << "GPU 运行开始:" << time_stamp() << endl;
error = clEnqueueNDRangeKernel(queue, run_as_gpu, 1, NULL, &global_work_size, NULL, 0, NULL, NULL);
cout << "GPU 运行结束:" << time_stamp() << endl;
OPENCL_CHECK_ERRORS(error) //取得kernel返回值
float* gpu_sum = new float[size];
clEnqueueReadBuffer(queue, sum_d, CL_TRUE, 0, mem_size, gpu_sum, 0, NULL, NULL); cout << "CPU 运行开始:" << time_stamp() << endl;
RunAsCpu(nums1_h, nums2_h, sum_h, size);
cout << "CPU 运行结束:" << time_stamp() << endl; assert(memcmp(sum_h, gpu_sum, size * sizeof(float)) == 0); delete[] nums1_h;
delete[] nums2_h;
delete[] sum_h;
delete[] gpu_sum;
delete[] platforms;
clReleaseKernel(run_as_gpu);
clReleaseCommandQueue(queue);
clReleaseContext(context);
clReleaseMemObject(nums1_d);
clReleaseMemObject(nums2_d);
clReleaseMemObject(sum_d);
return 0;

四、运行结果

由于运算比较简单,CPU和GPU几乎没差别,在后续复杂运算中应该是会有差别的。

OpenCL入门:(二:用GPU计算两个数组和)

五、相关下载

工程下载

六、后续

看了几篇文章后似乎简单使用OpenCL还是不复杂的,OpenCL关键应该在于如何优化性能,如何调用kernel函数,可以将GPU效果最优化。以后的文章一部分涉及OpenCL原理,一部分涉及到更复杂的运算,当然了,博主也是学习阶段,没有练手项目,只能从官方demos中找找了。

上一篇:PHP两个数组相加


下一篇:[LeetCode] Intersection of Two Arrays 两个数组相交