cuda理解并发内核执行

Question

我试图了解并发内核执行的工作原理.我写了一个简单的程序来试图理解它.内核将使用2个流填充2D数组.当有1个流,没有并发时,我得到了正确的结果.当我尝试使用2个流,尝试并发时,我得到了错误的结果.我相信它与内存传输有关,因为我不太确定我是否正确或我设置内核的方式.编程指南对我来说不够好.出于我的目的,我需要Matlab来调用内核.

据我了解,主程序将:

• 在主机上分配固定内存
• 分配单个流所需的GPU内存(2个流=主机总内存的一半)
• 创建流
• 循环流
• 使用cudaMemcpyAsync()将单个流的内存从主机复制到设备
• 为流执行内核
• 将流的内存复制回主机cudaMemcpyAsync()
  • 我相信我通过根据每个流的数据大小和流数量使用偏移量从每个流需要的位置引用内存来做正确的事情.
• 摧毁溪流
• 释放记忆

这是我试图使用的代码.

concurrentKernel.cpp

__global__ void concurrentKernel(int const width, 
                                  int const streamIdx,
                                  double *array)
 {
     int thread = (blockIdx.x * blockDim.x) + threadIdx.x;;

     for (int i = 0; i < width; i ++)
     {
        array[thread*width+i] = thread+i*width+1;
//         array[thread*width+i+streamIdx] = thread+i*width+streamIdx*width/2;
     }

 }

concurrentMexFunction.cu

#include <stdio.h>
#include <math.h>
#include "mex.h"

/* Kernel function */
#include "concurrentKernel.cpp"


void mexFunction(int        nlhs,
                 mxArray    *plhs[],
                 int        nrhs,
                 mxArray    *prhs[])
{

    int const numberOfStreams = 2; // set number of streams to use here.
    cudaError_t cudaError;
    int offset;

    int width, height, fullSize, streamSize;
    width = 512;
    height = 512;
    fullSize = height*width;
    streamSize = (int)(fullSize/numberOfStreams);
    mexPrintf("fullSize: %d, streamSize: %d\n",fullSize, streamSize);

    /* Return the populated array */
    double *returnedArray;
    plhs[0] = mxCreateDoubleMatrix(height, width, mxREAL);
    returnedArray = mxGetPr(plhs[0]);

    cudaStream_t stream[numberOfStreams];
    for (int i = 0; i < numberOfStreams; i++)
    {
        cudaStreamCreate(&stream[i]);    
    }

    /* host memory */
    double *hostArray;
    cudaError = cudaMallocHost(&hostArray,sizeof(double)*fullSize);    // full size of array.
    if (cudaError != cudaSuccess) {mexPrintf("hostArray memory allocation failed\n********** Error: %s **********\n",cudaGetErrorString(cudaError)); return; }

    for (int i = 0; i < height; i++)
    {
        for (int j = 0; j < width; j++)
        {
            hostArray[i*width+j] = -1.0;
        }
    }

    /* device memory */
    double *deviceArray;
    cudaError = cudaMalloc( (void **)&deviceArray,sizeof(double)*streamSize);    // size of array for each stream.
    if (cudaError != cudaSuccess) {mexPrintf("deviceArray memory allocation failed\n********** Error: %s **********\n",cudaGetErrorString(cudaError)); return; }


    for (int i = 0; i < numberOfStreams; i++)
    {
        offset = i;//*streamSize;
        mexPrintf("offset: %d, element: %d\n",offset*sizeof(double),offset);

        cudaMemcpyAsync(deviceArray, hostArray+offset, sizeof(double)*streamSize, cudaMemcpyHostToDevice, stream[i]);
        if (cudaError != cudaSuccess) {mexPrintf("deviceArray memory allocation failed\n********** Error: %s **********\n",cudaGetErrorString(cudaError)); return; }

        concurrentKernel<<<1, 512, 0, stream[i]>>>(width, i, deviceArray);

        cudaMemcpyAsync(returnedArray+offset, deviceArray, sizeof(double)*streamSize, cudaMemcpyDeviceToHost, stream[i]);
        if (cudaError != cudaSuccess) {mexPrintf("returnedArray memory allocation failed\n********** Error: %s **********\n",cudaGetErrorString(cudaError)); return; }

        mexPrintf("returnedArray[offset]: %g, [end]: %g\n",returnedArray[offset/sizeof(double)],returnedArray[(i+1)*streamSize-1]);
    }


    for (int i = 0; i < numberOfStreams; i++)
    {
        cudaStreamDestroy(stream[i]);    
    }

    cudaFree(hostArray);
    cudaFree(deviceArray);

}

当有2个流时,结果是一个零数组,这让我觉得它我的内存有问题.谁能解释我做错了什么？如果有人需要帮助从Matlab编译和运行这些,我可以提供命令来执行此操作.

更新:

for (int i = 0; i < numberOfStreams; i++)
{
    offset = i*streamSize;
    mexPrintf("offset: %d, element: %d\n",offset*sizeof(double),offset);

    cudaMemcpyAsync(deviceArray, hostArray+offset, sizeof(double)*streamSize, cudaMemcpyHostToDevice, stream[i]);
    if (cudaError != cudaSuccess) {mexPrintf("deviceArray memory allocation failed\n********** Error: %s **********\n",cudaGetErrorString(cudaError)); return; }

    concurrentKernel<<<1, 512, 0, stream[i]>>>(width, i, deviceArray);


}
cudaDeviceSynchronize();


for (int i = 0; i < numberOfStreams; i++)
{
    offset = i*streamSize;
    mexPrintf("offset: %d, element: %d\n",offset*sizeof(double),offset);

    cudaMemcpyAsync(returnedArray+offset, deviceArray, sizeof(double)*streamSize, cudaMemcpyDeviceToHost, stream[i]);
    if (cudaError != cudaSuccess) {mexPrintf("returnedArray memory allocation failed\n********** Error: %s **********\n",cudaGetErrorString(cudaError)); return; }

    mexPrintf("returnedArray[offset]: %g, [end]: %g\n",returnedArray[offset/sizeof(double)],returnedArray[(i+1)*streamSize-1]);

    cudaStreamDestroy(stream[i]);    
}

Answer 1

您需要记住,与流一起使用的API是完全异步的,因此控制会立即返回到调用主机线程.如果您没有在运行异步操作的GPU和主机之间插入某种同步点,则无法保证您在流中排队的操作实际上已完成.在你的例子中,这意味着需要这样的东西:

for (int i = 0; i < numberOfStreams; i++) 
{ 
    offset = i;//*streamSize; 
    mexPrintf("offset: %d, element: %d\n",offset*sizeof(double),offset); 

    cudaMemcpyAsync(deviceArray, hostArray+offset, sizeof(double)*streamSize, 
                    cudaMemcpyHostToDevice, stream[i]); 

    concurrentKernel<<<1, 512, 0, stream[i]>>>(width, i, deviceArray); 

    cudaMemcpyAsync(returnedArray+offset, deviceArray, sizeof(double)*streamSize,
                    cudaMemcpyDeviceToHost, stream[i]); 
} 

// Host thread waits here until both kernels and copies are finished
cudaDeviceSynchronize();

for (int i = 0; i < numberOfStreams; i++) 
{ 
    mexPrintf("returnedArray[offset]: %g, [end]: %g\n",returnedArray[offset/sizeof(double)],returnedArray[(i+1)*streamSize-1]); 
    cudaStreamDestroy(stream[i]);     
} 

这里的关键是在尝试检查主机内存中的结果之前,需要确保两个内存传输都已完成.您的原始代码和更新都不会这样做.