我正在尝试使用JCuda中的共享内存实现简单的矩阵乘法程序。
以下是我的JCudaSharedMatrixMul.java代码：
import static jcuda.driver.JCudaDriver.cuCtxCreate;
import static jcuda.driver.JCudaDriver.cuCtxSynchronize;
import static jcuda.driver.JCudaDriver.cuDeviceGet;
import static jcuda.driver.JCudaDriver.cuInit;
import static jcuda.driver.JCudaDriver.cuLaunchKernel;
import static jcuda.driver.JCudaDriver.cuMemAlloc;
import static jcuda.driver.JCudaDriver.cuMemFree;
import static jcuda.driver.JCudaDriver.cuMemcpyDtoH;
import static jcuda.driver.JCudaDriver.cuMemcpyHtoD;
import static jcuda.driver.JCudaDriver.cuModuleGetFunction;
import static jcuda.driver.JCudaDriver.cuModuleLoad;
import static jcuda.runtime.JCuda.cudaEventCreate;
import static jcuda.runtime.JCuda.cudaEventRecord;
import static jcuda.runtime.JCuda.*;
import java.io.ByteArrayOutputStream;
import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import java.util.Scanner;
import jcuda.Pointer;
import jcuda.Sizeof;
import jcuda.driver.CUcontext;
import jcuda.driver.CUdevice;
import jcuda.driver.CUdeviceptr;
import jcuda.driver.CUfunction;
import jcuda.driver.CUmodule;
import jcuda.driver.JCudaDriver;
import jcuda.runtime.cudaEvent_t;
public class JCudaSharedMatrixMul
{
public static void main(String[] args) throws IOException
{
// Enable exceptions and omit all subsequent error checks
JCudaDriver.setExceptionsEnabled(true);
// Create the PTX file by calling the NVCC
String ptxFilename = preparePtxFile("JCudaSharedMatrixMulKernel.cu");
//Initialize the driver and create a context for the first device.
cuInit(0);
CUdevice device = new CUdevice();
cuDeviceGet (device, 0);
CUcontext context = new CUcontext();
cuCtxCreate(context, 0, device);
//Load PTX file
CUmodule module = new CUmodule();
cuModuleLoad(module,ptxFilename);
//Obtain a function pointer to the Add function
CUfunction function = new CUfunction();
cuModuleGetFunction(function, module, "jCudaSharedMatrixMulKernel");
int numRows = 16;
int numCols = 16;
//Allocate and fill Host input Matrices:
float hostMatrixA[] = new float[numRows*numCols];
float hostMatrixB[] = new float[numRows*numCols];
float hostMatrixC[] = new float[numRows*numCols];
for(int i = 0; i<numRows; i++)
{
for(int j = 0; j<numCols; j++)
{
hostMatrixA[i*numCols+j] = (float) 1;
hostMatrixB[i*numCols+j] = (float) 1;
}
}
// Allocate the device input data, and copy the
// host input data to the device
CUdeviceptr devMatrixA = new CUdeviceptr();
cuMemAlloc(devMatrixA, numRows * numCols * Sizeof.FLOAT);
//This is the part where it gives me the error
cuMemcpyHtoD(devMatrixA, Pointer.to(hostMatrixA), numRows * numCols * Sizeof.FLOAT);
CUdeviceptr devMatrixB = new CUdeviceptr();
cuMemAlloc(devMatrixB, numRows * numCols * Sizeof.FLOAT);
//This is the part where it gives me the error
cuMemcpyHtoD(devMatrixB, Pointer.to(hostMatrixB ), numRows * numCols * Sizeof.FLOAT);
//Allocate device matrix C to store output
CUdeviceptr devMatrixC = new CUdeviceptr();
cuMemAlloc(devMatrixC, numRows * numCols * Sizeof.FLOAT);
// Set up the kernel parameters: A pointer to an array
// of pointers which point to the actual values.
Pointer kernelParameters = Pointer.to(
Pointer.to(new int[]{numCols}),
Pointer.to(devMatrixA),
Pointer.to(devMatrixB),
Pointer.to(devMatrixC));
//Kernel thread configuration
int blockSize = 16;
int gridSize = 1;
cudaEvent_t start = new cudaEvent_t();
cudaEvent_t stop = new cudaEvent_t();
cudaEventCreate(start);
cudaEventCreate(stop);
long start_nano=System.nanoTime();
cudaEventRecord(start, null);
cuLaunchKernel(function,
gridSize, 1, 1,
blockSize, 16, 1,
250, null, kernelParameters, null);
cuCtxSynchronize();
cudaEventRecord(stop, null);
long end_nano=System.nanoTime();
float elapsedTimeMsArray[] = { Float.NaN };
cudaEventElapsedTime(elapsedTimeMsArray, start, stop);
float elapsedTimeMs = elapsedTimeMsArray[0];
System.out.println("Time Required (Using cudaevent elapsed time) = " + " " +elapsedTimeMs+
"Time Required (Using nanotime)= "+(end_nano-start_nano)/1000000);
// Allocate host output memory and copy the device output
// to the host.
//This is the part where it gives me the error
cuMemcpyDtoH(Pointer.to(hostMatrixC), devMatrixC, numRows * numCols * Sizeof.FLOAT);
//verify the result
for (int i =0; i<numRows; i++)
{
for (int j =0; j<numRows; j++)
{
System.out.print(" "+ hostMatrixC[i*numCols+j]);
}
System.out.println("");
}
cuMemFree(devMatrixA);
cuMemFree(devMatrixB);
cuMemFree(devMatrixC);
}
private static String preparePtxFile(String cuFileName) throws IOException
{
int endIndex = cuFileName.lastIndexOf('.');
if (endIndex == -1)
endIndex = cuFileName.length()-1;
{
}
String ptxFileName = cuFileName.substring(0, endIndex+1)+"ptx";
File ptxFile = new File(ptxFileName);
if (ptxFile.exists())
{
return ptxFileName;
}
File cuFile = new File(cuFileName);
if (!cuFile.exists())
{
throw new IOException("Input file not found: "+cuFileName);
}
String modelString = "-m"+System.getProperty("sun.arch.data.model");
String command = "nvcc " + modelString + " -ptx "+ cuFile.getPath()+" -o "+ptxFileName;
System.out.println("Executing\n"+command);
Process process = Runtime.getRuntime().exec(command);
String errorMessage = new String(toByteArray(process.getErrorStream()));
String outputMessage = new String(toByteArray(process.getInputStream()));
int exitValue = 0;
try
{
exitValue = process.waitFor();
}
catch (InterruptedException e)
{
Thread.currentThread().interrupt();
throw new IOException(
"Interrupted while waiting for nvcc output", e);
}
if (exitValue != 0)
{
System.out.println("nvcc process exitValue "+exitValue);
System.out.println("errorMessage:\n"+errorMessage);
System.out.println("outputMessage:\n"+outputMessage);
throw new IOException(
"Could not create .ptx file: "+errorMessage);
}
System.out.println("Finished creating PTX file");
return ptxFileName;
}
private static byte[] toByteArray(InputStream inputStream) throws IOException
{
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte buffer[] = new byte[8192];
while (true)
{
int read = inputStream.read(buffer);
if (read == -1)
{
break;
}
baos.write(buffer, 0, read);
}
return baos.toByteArray();
}
}
以下是我的JCudaSharedMatrixMulKernel.cu代码：
extern "C"
__global__ void jCudaSharedMatrixMulKernel(int N,float *ad,float *bd,float *cd)
{
float pvalue=0;
int TILE=blockDim.x;
int ty=threadIdx.y;
int tx=threadIdx.x;
__shared__ float ads[4][4];
__shared__ float bds[4][4];
int Row = blockIdx.y * blockDim.y + threadIdx.y;
int Col = blockIdx.x * blockDim.x + threadIdx.x;
for(int i=0;i< N/TILE;++i)
{
ads[ty][tx] = ad[Row * N + (i * TILE) + tx];
bds[ty][tx] = bd[(i * TILE + ty) * N + Col];
__syncthreads();
for(int k=0;k<TILE;k++)
pvalue += ads[ty][k] * bds[k][tx];
__syncthreads();
}
cd[Row * N + Col] = pvalue;
}
在上面的示例中，每个块使用的总共享内存是2 * 4 * 4 * 4 = 128个字节。 在cuLaunchKernel中，当我将 sharedMemBytes 参数定义为0（零）时，它会给我以下错误：
**Exception in thread "main" jcuda.CudaException: CUDA_ERROR_LAUNCH_FAILED
at jcuda.driver.JCudaDriver.checkResult(JCudaDriver.java:282)
at jcuda.driver.JCudaDriver.cuCtxSynchronize(JCudaDriver.java:1795)
at JCudaSharedMatrixMul.main(JCudaSharedMatrixMul.java:121)**
当我将其定义为128时，它会给出相同的上述错误。但是当我把它设为129然后它给了我正确的输出！当我给出介于129到49024之间的任何值时，它会给出正确的结果。 我的问题是，当我将其定义为128时，为什么我无法获得正确的输出？还可以定义什么是最大共享内存？为什么这个129-49024系列在这里工作？
答案 0 :(得分：3)
您正在启动16x16线程的块：
cuLaunchKernel(function,
gridSize, 1, 1,
blockSize, 16, 1, <-- the first two params are block.x and block.y
250, null, kernelParameters, null);
所以__shared__ float ads[4][4];根本不应该工作。例如，这些内核代码行将访问某些线程的那些共享数组越界：
ads[ty][tx] = ad[Row * N + (i * TILE) + tx];
bds[ty][tx] = bd[(i * TILE + ty) * N + Col];
^ ^
| tx goes from 0..15 for a 16x16 threadblock
ty goes from 0..15 for a 16x16 threadblock
你的代码在这方面被打破了。如果您使用cuda-memcheck运行代码，它可能会捕获这些越界访问，即使在您的＆＃34;传递＆＃34;案件。查看matrixMulDrv cuda sample code将具有指导意义，并且您会看到共享内存分配为2*block_size*block_size，因为它也适用于您的情况，但您的共享内存定义应该是[16][16]而不是[4][4]当您超过128个字节时，共享内存分配粒度可能会正常工作，但代码中存在缺陷。
您的共享定义应为：
__shared__ float ads[16][16];
__shared__ float bds[16][16];
由于上述分配是静态分配，sharedMemBytes参数是defined as动态共享内存分配，因此对于此示例，您不需要分配任何（0为OK）动态共享记忆，它仍然有效。静态和动态之间的区别包括here。
每个块的最大共享内存可在the documentation中使用，或者如果您运行cuda deviceQuery示例代码。对于cc2.0和更新的设备，它是48K字节。