OpenCL中矩阵乘法中的右global_work_size和local_work_size是什么？
What is the right global_work_size and local_work_size in Matrix Multiplication in OpenCL?
我在 JOCL 中做矩阵乘法 Java 绑定到 OpenCL。问题，我得到一个错误
Exception in thread "main" org.jocl.CLException: CL_INVALID_WORK_ITEM_SIZE
at org.jocl.CL.checkResult(CL.java:787)
at org.jocl.CL.clEnqueueNDRangeKernel(CL.java:20802)
at org.jocl.samples.JOCLSample.main(JOCLSample.java:150)
代码如下：
package org.jocl.samples;
import static org.jocl.CL.*;
import org.jocl.*;
import static java.lang.System.nanoTime;
public class JOCLSample
{
/**
* The source code of the OpenCL program to execute
*/
private static String programSource =
"__kernel void "+
"matrixMul(__global float* C,"+
" __global float* A,"+
" __global float* B,"+
" int wA, int wB)"+
"{"+
"int tx = get_global_id(0);"+
"int ty = get_global_id(1);"+
"float value = 0;"+
"for (int k = 0; k < wA; ++k)"+
"{"+
" float elementA = A[ty * wA + k];"+
" float elementB = B[k * wB + tx];"+
" value += elementA * elementB;"+
"}"+
"C[ty * wA + tx] = value;"+
"}";
/**
* The entry point of this sample
*
* @param args Not used
*/
public static void main(String args[])
{
// Create input- and output data
int n = 10;
float srcArrayA[] = new float[n];
float srcArrayB[] = new float[n];
float dstArray[] = new float[n];
for (int i=0; i<n; i++)
{
srcArrayA[i] = i;
srcArrayB[i] = i;
}
Pointer srcA = Pointer.to(srcArrayA);
Pointer srcB = Pointer.to(srcArrayB);
Pointer dst = Pointer.to(dstArray);
// The platform, device type and device number
// that will be used
final int platformIndex = 0;
final long deviceType = CL_DEVICE_TYPE_ALL;
final int deviceIndex = 0;
// Enable exceptions and subsequently omit error checks in this sample
CL.setExceptionsEnabled(true);
// Obtain the number of platforms
int numPlatformsArray[] = new int[1];
clGetPlatformIDs(0, null, numPlatformsArray);
int numPlatforms = numPlatformsArray[0];
// Obtain a platform ID
cl_platform_id platforms[] = new cl_platform_id[numPlatforms];
clGetPlatformIDs(platforms.length, platforms, null);
cl_platform_id platform = platforms[platformIndex];
// Initialize the context properties
cl_context_properties contextProperties = new cl_context_properties();
contextProperties.addProperty(CL_CONTEXT_PLATFORM, platform);
// Obtain the number of devices for the platform
int numDevicesArray[] = new int[1];
clGetDeviceIDs(platform, deviceType, 0, null, numDevicesArray);
int numDevices = numDevicesArray[0];
// Obtain a device ID
cl_device_id devices[] = new cl_device_id[numDevices];
clGetDeviceIDs(platform, deviceType, numDevices, devices, null);
cl_device_id device = devices[deviceIndex];
// Create a context for the selected device
cl_context context = clCreateContext(
contextProperties, 1, new cl_device_id[]{device},
null, null, null);
// Create a command-queue for the selected device
cl_command_queue commandQueue =
clCreateCommandQueue(context, device, 0, null);
// Allocate the memory objects for the input- and output data
cl_mem memObjects[] = new cl_mem[3];
memObjects[0] = clCreateBuffer(context,
CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
Sizeof.cl_float * n, srcA, null);
memObjects[1] = clCreateBuffer(context,
CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
Sizeof.cl_float * n, srcB, null);
memObjects[2] = clCreateBuffer(context,
CL_MEM_READ_WRITE,
Sizeof.cl_float * n, null, null);
// Create the program from the source code
cl_program program = clCreateProgramWithSource(context,
1, new String[]{ programSource }, null, null);
// Build the program
clBuildProgram(program, 0, null, null, null, null);
// Create the kernel
cl_kernel kernel = clCreateKernel(program, "matrixMul", null);
long time = nanoTime();
// Set the arguments for the kernel
clSetKernelArg(kernel, 0,
Sizeof.cl_mem, Pointer.to(memObjects[0]));
clSetKernelArg(kernel, 1,
Sizeof.cl_mem, Pointer.to(memObjects[1]));
clSetKernelArg(kernel, 2,
Sizeof.cl_mem, Pointer.to(memObjects[2]));
// Set the work-item dimensions
long global_work_size[] = new long[]{n};
long local_work_size[] = new long[]{1};
// Execute the kernel
clEnqueueNDRangeKernel(commandQueue, kernel, 2, null,
global_work_size, local_work_size, 0, null, null);
// Read the output data
clEnqueueReadBuffer(commandQueue, memObjects[2], CL_TRUE, 0,
n * Sizeof.cl_float, dst, 0, null, null);
time = nanoTime() - time;
System.out.println("GPU time: "+ time +"ns " + (time/1000000)+"ms");
// Release kernel, program, and memory objects
clReleaseMemObject(memObjects[0]);
clReleaseMemObject(memObjects[1]);
clReleaseMemObject(memObjects[2]);
clReleaseKernel(kernel);
clReleaseProgram(program);
clReleaseCommandQueue(commandQueue);
clReleaseContext(context);
// Verify the result
boolean passed = true;
final float epsilon = 1e-7f;
for (int i=0; i<n; i++)
{
float x = dstArray[i];
float y = srcArrayA[i] * srcArrayB[i];
boolean epsilonEqual = Math.abs(x - y) <= epsilon * Math.abs(x);
if (!epsilonEqual)
{
passed = false;
break;
}
}
System.out.println("Test "+(passed?"PASSED":"FAILED"));
if (n <= 128)
{
System.out.println("Result: "+java.util.Arrays.toString(dstArray));
}
}
}
我的问题是，大小为[1024 x 1024]、[2048 x 2048]、[4096 x 4096]和[的矩阵乘法的右global_work_size和local_work_size是多少？ 8192 x 8192]？
这是 global_work_size 和 local_work_size 中导致错误的代码
// Set the work-item dimensions
long global_work_size[] = new long[]{n};
long local_work_size[] = new long[]{1};
// Execute the kernel
clEnqueueNDRangeKernel(commandQueue, kernel, 2, null,
global_work_size, local_work_size, 0, null, null);
除非您正在使用共享本地内存、async_copy 函数，或出于其他原因检查 get_local_id，否则您不需要设置本地工作组大小，您可以让driver select 一个（通过传递 NULL）。一旦一切正常，您可以尝试传递不同的值以查看它是否会改变性能，但请注意全局大小需要是局部大小的整数倍，因此如果这会使它大于您的实际工作大小，您必须在内核中添加条件以跳过边界情况。
您正在将 clEnqueueNDRange 中的 work_dim 参数分配给 2。但是您传递的全局大小数组的大小为 1（您使用单个值对其进行初始化）。这是轰炸，因为它希望数组有 2 个条目。
work_dim 字段指定全局工作项的维度，而不是数据数组的维度。所以将 work_dim 字段（第三个参数）设置为 1.
我在 JOCL 中做矩阵乘法 Java 绑定到 OpenCL。问题，我得到一个错误
Exception in thread "main" org.jocl.CLException: CL_INVALID_WORK_ITEM_SIZE
at org.jocl.CL.checkResult(CL.java:787)
at org.jocl.CL.clEnqueueNDRangeKernel(CL.java:20802)
at org.jocl.samples.JOCLSample.main(JOCLSample.java:150)
代码如下：
package org.jocl.samples;
import static org.jocl.CL.*;
import org.jocl.*;
import static java.lang.System.nanoTime;
public class JOCLSample
{
/**
* The source code of the OpenCL program to execute
*/
private static String programSource =
"__kernel void "+
"matrixMul(__global float* C,"+
" __global float* A,"+
" __global float* B,"+
" int wA, int wB)"+
"{"+
"int tx = get_global_id(0);"+
"int ty = get_global_id(1);"+
"float value = 0;"+
"for (int k = 0; k < wA; ++k)"+
"{"+
" float elementA = A[ty * wA + k];"+
" float elementB = B[k * wB + tx];"+
" value += elementA * elementB;"+
"}"+
"C[ty * wA + tx] = value;"+
"}";
/**
* The entry point of this sample
*
* @param args Not used
*/
public static void main(String args[])
{
// Create input- and output data
int n = 10;
float srcArrayA[] = new float[n];
float srcArrayB[] = new float[n];
float dstArray[] = new float[n];
for (int i=0; i<n; i++)
{
srcArrayA[i] = i;
srcArrayB[i] = i;
}
Pointer srcA = Pointer.to(srcArrayA);
Pointer srcB = Pointer.to(srcArrayB);
Pointer dst = Pointer.to(dstArray);
// The platform, device type and device number
// that will be used
final int platformIndex = 0;
final long deviceType = CL_DEVICE_TYPE_ALL;
final int deviceIndex = 0;
// Enable exceptions and subsequently omit error checks in this sample
CL.setExceptionsEnabled(true);
// Obtain the number of platforms
int numPlatformsArray[] = new int[1];
clGetPlatformIDs(0, null, numPlatformsArray);
int numPlatforms = numPlatformsArray[0];
// Obtain a platform ID
cl_platform_id platforms[] = new cl_platform_id[numPlatforms];
clGetPlatformIDs(platforms.length, platforms, null);
cl_platform_id platform = platforms[platformIndex];
// Initialize the context properties
cl_context_properties contextProperties = new cl_context_properties();
contextProperties.addProperty(CL_CONTEXT_PLATFORM, platform);
// Obtain the number of devices for the platform
int numDevicesArray[] = new int[1];
clGetDeviceIDs(platform, deviceType, 0, null, numDevicesArray);
int numDevices = numDevicesArray[0];
// Obtain a device ID
cl_device_id devices[] = new cl_device_id[numDevices];
clGetDeviceIDs(platform, deviceType, numDevices, devices, null);
cl_device_id device = devices[deviceIndex];
// Create a context for the selected device
cl_context context = clCreateContext(
contextProperties, 1, new cl_device_id[]{device},
null, null, null);
// Create a command-queue for the selected device
cl_command_queue commandQueue =
clCreateCommandQueue(context, device, 0, null);
// Allocate the memory objects for the input- and output data
cl_mem memObjects[] = new cl_mem[3];
memObjects[0] = clCreateBuffer(context,
CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
Sizeof.cl_float * n, srcA, null);
memObjects[1] = clCreateBuffer(context,
CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
Sizeof.cl_float * n, srcB, null);
memObjects[2] = clCreateBuffer(context,
CL_MEM_READ_WRITE,
Sizeof.cl_float * n, null, null);
// Create the program from the source code
cl_program program = clCreateProgramWithSource(context,
1, new String[]{ programSource }, null, null);
// Build the program
clBuildProgram(program, 0, null, null, null, null);
// Create the kernel
cl_kernel kernel = clCreateKernel(program, "matrixMul", null);
long time = nanoTime();
// Set the arguments for the kernel
clSetKernelArg(kernel, 0,
Sizeof.cl_mem, Pointer.to(memObjects[0]));
clSetKernelArg(kernel, 1,
Sizeof.cl_mem, Pointer.to(memObjects[1]));
clSetKernelArg(kernel, 2,
Sizeof.cl_mem, Pointer.to(memObjects[2]));
// Set the work-item dimensions
long global_work_size[] = new long[]{n};
long local_work_size[] = new long[]{1};
// Execute the kernel
clEnqueueNDRangeKernel(commandQueue, kernel, 2, null,
global_work_size, local_work_size, 0, null, null);
// Read the output data
clEnqueueReadBuffer(commandQueue, memObjects[2], CL_TRUE, 0,
n * Sizeof.cl_float, dst, 0, null, null);
time = nanoTime() - time;
System.out.println("GPU time: "+ time +"ns " + (time/1000000)+"ms");
// Release kernel, program, and memory objects
clReleaseMemObject(memObjects[0]);
clReleaseMemObject(memObjects[1]);
clReleaseMemObject(memObjects[2]);
clReleaseKernel(kernel);
clReleaseProgram(program);
clReleaseCommandQueue(commandQueue);
clReleaseContext(context);
// Verify the result
boolean passed = true;
final float epsilon = 1e-7f;
for (int i=0; i<n; i++)
{
float x = dstArray[i];
float y = srcArrayA[i] * srcArrayB[i];
boolean epsilonEqual = Math.abs(x - y) <= epsilon * Math.abs(x);
if (!epsilonEqual)
{
passed = false;
break;
}
}
System.out.println("Test "+(passed?"PASSED":"FAILED"));
if (n <= 128)
{
System.out.println("Result: "+java.util.Arrays.toString(dstArray));
}
}
}
我的问题是，大小为[1024 x 1024]、[2048 x 2048]、[4096 x 4096]和[的矩阵乘法的右global_work_size和local_work_size是多少？ 8192 x 8192]？
这是 global_work_size 和 local_work_size 中导致错误的代码
// Set the work-item dimensions
long global_work_size[] = new long[]{n};
long local_work_size[] = new long[]{1};
// Execute the kernel
clEnqueueNDRangeKernel(commandQueue, kernel, 2, null,
global_work_size, local_work_size, 0, null, null);
除非您正在使用共享本地内存、async_copy 函数，或出于其他原因检查 get_local_id，否则您不需要设置本地工作组大小，您可以让driver select 一个（通过传递 NULL）。一旦一切正常，您可以尝试传递不同的值以查看它是否会改变性能，但请注意全局大小需要是局部大小的整数倍，因此如果这会使它大于您的实际工作大小，您必须在内核中添加条件以跳过边界情况。
您正在将 clEnqueueNDRange 中的 work_dim 参数分配给 2。但是您传递的全局大小数组的大小为 1（您使用单个值对其进行初始化）。这是轰炸，因为它希望数组有 2 个条目。 work_dim 字段指定全局工作项的维度，而不是数据数组的维度。所以将 work_dim 字段（第三个参数）设置为 1.