我制作了一个OpenCL程序并使用固定内存(CL_MEM_ALLOC_HOST_PTR
)来提高设备之间的传输速率。
传输速率按我的预期增加(使用AMD APP Profiler 2.4获得传输速率)。 问题是矩阵4096 x 4096(64 MB)的传输速率高于PCIe带宽(93703 GB / s)。
当我使用零拷贝缓冲区(CL_MEM_ALLOC_HOST_PTR + clEnqueueMapBuffer)时也发生了这种情况。 我搜索一些信息,如果固定内存和零复制缓冲区具有高传输速率,但它仍然受限于PCIe带宽用于独立GPU。 那么,如果传输速率超过PCIe带宽(使用PCIe带宽2.0 x 16),这是正常的吗?
我的操作系统是Windows 7 64位。 我使用AMD APP SDK 2.6和独立GPU AMD HD 6630M。
编辑: 这是代码:
#include <Windows.h>
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
#ifdef __APPLE__
#include <OpenCL/opencl.h>
#else
#include <CL/cl.h>
#endif
#define MAX_SOURCE_SIZE (0x100000)
cl_context context = NULL;
cl_command_queue queue = NULL;
cl_program program = NULL;
void MatrixMul(cl_mem d_A, cl_mem d_B, cl_mem d_C, int size)
{
cl_int err;
cl_kernel naive;
// Create Kernel Object Bound To Kernel Function
naive = clCreateKernel(program, "naiveAlgorithm", &err);
//Set size of global work item and work tem in each work goups
int globalsize = size;
int localsize;
if(globalsize >= 16)
{
localsize =16;
}else
{
localsize = globalsize;
}
size_t global_work_items [2] = {globalsize, globalsize};
size_t local_work_items [2] = {localsize, localsize};
// Setup Kernel Argument
err = clSetKernelArg(naive, 0, sizeof(cl_mem), (void *)&d_A);
err = clSetKernelArg(naive, 1, sizeof(cl_mem), (void *)&d_B);
err = clSetKernelArg(naive, 2, sizeof(cl_mem), (void *)&d_C);
err = clSetKernelArg(naive, 3, sizeof(cl_int), (void *)&size);
// Execute OpenCL kernel for Naive Algorithm
err = clEnqueueNDRangeKernel(queue, naive, 2, NULL, global_work_items, local_work_items, 0, NULL, NULL);
clFinish(queue);
//Release Kernel
err = clReleaseKernel(naive);
}
void Naive(cl_float* matrixA, cl_float* matrixB, cl_float* matrixC, int size)
{
int err;
// OpenCL device memory for matrices
cl_mem d_A;
cl_mem d_B;
cl_mem d_C;
// Allocate Device Memory For Input And Output
d_A = clCreateBuffer(context, CL_MEM_READ_ONLY , sizeof(cl_float)*size*size, 0, &err);
d_B = clCreateBuffer(context, CL_MEM_READ_ONLY , sizeof(cl_float)*size*size, 0, &err);
d_C = clCreateBuffer(context, CL_MEM_WRITE_ONLY|CL_MEM_ALLOC_HOST_PTR ,sizeof(cl_float)*size*size, 0,&err);
// Copy Host Memory To Memory Device
err = clEnqueueWriteBuffer(queue, d_A, CL_FALSE, 0, sizeof(cl_float)*size*size, matrixA, 0, NULL, NULL);
err = clEnqueueWriteBuffer(queue, d_B, CL_FALSE, 0, sizeof(cl_float)*size*size, matrixB, 0, NULL, NULL);
MatrixMul(d_A, d_B, d_C, size);
err = clEnqueueReadBuffer(queue, d_C, CL_TRUE, 0, sizeof(cl_float)*size*size, matrixC, 0, NULL, NULL);
err = clReleaseMemObject(d_A);
err = clReleaseMemObject(d_B);
err = clReleaseMemObject(d_C);
}
//Main Function
int main(int argc, char **argv)
{
//Size of matrix for Strassen Algorithm
cl_int size = 4096;
//Matrix for input and output
cl_float * matrixA;
cl_float * matrixB;
cl_float * matrixC;
//Allocate and init memory for the host
matrixA = (cl_float *) malloc(size*size*sizeof(cl_float));
matrixB = (cl_float *) malloc(size*size*sizeof(cl_float));
matrixC = (cl_float *) malloc(size*size*sizeof(cl_float));
//Fill matrix
fillMatrix(matrixA,size);
fillMatrix(matrixB,size);
//print input for matrix A and B
cout<<"Input for matrix A :"<<endl;
printMatrix(matrixA, size*size, size);
cout<<"Input for matrix B :"<<endl;
printMatrix(matrixB, size*size, size);
cl_int err; // error code
cl_platform_id* platforms;
cl_uint platformCount;
cl_device_id device;
int platformtype = 0; //if 0 using amd app sdk but if 1 using intel sdk
clGetPlatformIDs(0, NULL, &platformCount); //get number of platform
platforms = (cl_platform_id*) malloc(sizeof(cl_platform_id) * platformCount);
clGetPlatformIDs(platformCount, platforms, NULL); //get list of platform
clGetDeviceIDs (platforms [platformtype], CL_DEVICE_TYPE_GPU, 1, &device, NULL); //get list of devices
const cl_context_properties contextProperties [] =
{CL_CONTEXT_PLATFORM,
reinterpret_cast<cl_context_properties> (platforms [platformtype]),
0, 0
};
context = clCreateContext(contextProperties, 1, &device, NULL, NULL, &err);
![enter image description here][2]queue = clCreateCommandQueue(context, device, CL_QUEUE_PROFILING_ENABLE, &err);
//Load Kernel Source
FILE *fp;
const char fileName[] = "./MatMul_Kernel.cl";
size_t source_size;
char *source_str;
fp = fopen(fileName, "r");
if (!fp)
{
fprintf(stderr, "Failed to load kernel.\n");
exit(1);
}
source_str = (char *)malloc(MAX_SOURCE_SIZE);
source_size = fread(source_str, 1, MAX_SOURCE_SIZE, fp);
fclose(fp);
// Create Program Object
program = clCreateProgramWithSource(context, 1, (const char **) &source_str,(const size_t *),
&source_size, &err);
// Build Program
err = clBuildProgram(program, 1, &device, NULL, NULL, NULL);
Naive(matrixA, matrixB, matrixC, size);
//Cleanup all memory
err = clFlush(queue);
err = clFinish(queue);
err = clReleaseProgram(program);
err = clReleaseCommandQueue(queue);
err = clReleaseContext(context);
// Display result of matrix multiplication
cout<<"Output for matrix C :"<<endl;
printMatrix(matrixC, size*size, size);
cout<<endl;
free(matrixA);
free(matrixB);
free(matrixC);
free(source_str);
return 0;
}
这是内核代码:
__kernel void naiveAlgorithm(__global float *A, __global float *B, __global float *C, int size) {
int tx = get_global_id(0); //2D Thread IDx
int ty = get_global_id(1); //2D Thread IDy
float sum = 0;
//Calculate result of one element of Matrix C
for (int k = 0; k < size; k++) {
sum += A[ty*size+k] * B[k*size+tx];
}
C[ty*size+tx] = sum;
}
这是图像:
答案 0 :(得分:1)
由于以下行中的CL_MEM_ALLOC_HOST_PTR
标志,我看到您的输出数组实际上位于主机内存中:
d_C = clCreateBuffer(context, CL_MEM_WRITE_ONLY|CL_MEM_ALLOC_HOST_PTR ,sizeof(cl_float)*size*size, 0,&err);
这意味着您应该使用clEnqueueMapBuffer
,然后以您认为合适的方式使用矩阵,然后使用clEnqueueUnmapMemObject
。由于d_C已经在主机存储器中,所以不需要阵列matrixC。
在内核运行时,实际上会发生从GPU到主机的数据传输。地图调用确保所有数据已完成从GPU移动到CPU。这就是为什么传输时间实际上很小。
我无法找到有关clEnqueueReadBuffer
是否适用于固定内存的任何文档。我还看到您正在检索每个操作的错误代码,但不检查这些错误代码,因此您的代码可能会无声地失败。
关于clEnqueueReadBuffer
所花费的时间与传输数据所花费的时间之间的巨大差异,请注意所有排队的操作都不会立即发送到GPU。延迟的一个来源是用于图形卡的Windows显示驱动程序模型(WDDM)。用于clEnqueueReadBuffer
的+ -20微秒听起来适合这种延迟(我实际上看到了更长的延迟)。