Question

我已加载大小为1080 x 1920的图片（8位，无符号字符）。出于测试目的，我使用for loop处理相同的图像4次，然后生成其时间线分析。

策略：我将图片分成3部分。我已经制作了三个流来处理整个图像。

我在下面提供了一个最小的工作示例。我很抱歉它需要使用OpenCV的图像，但我不知道如何在不使用OpenCV加载图像的情况下模仿相同的情况。

问题：时间线分析显示第一个流已完成数据传输，但分配给它的内核仍未启动。分配给第一个流的内核和第三个流的数据传输是并行的。 所以，我的问题是为什么第一个流内核的处理没有与第二个流的数据传输并行启动？

GPU： NVIDIA Quadro K2000，兼容3.0

时间轴配置文件：为每个流分配了不同的颜色。

我的代码：

__global__ void multiStream_ColorTransformation_kernel(int numChannels, int iw, int ih, unsigned char *ptr_source, unsigned char *ptr_dst)
{
    // Calculate our pixel's location
    int x = (blockIdx.x * blockDim.x) + threadIdx.x;
    int y = (blockIdx.y * blockDim.y) + threadIdx.y;

    // Operate only if we are in the correct boundaries
    if (x >= 0 && x < iw && y >= 0 && y < ih / 3)
    {
        ptr_dst[numChannels*  (iw*y + x) + 0] = ptr_source[numChannels*  (iw*y + x) + 0];
        ptr_dst[numChannels*  (iw*y + x) + 1] = ptr_source[numChannels*  (iw*y + x) + 1];
        ptr_dst[numChannels*  (iw*y + x) + 2] = ptr_source[numChannels*  (iw*y + x) + 2];

    }
}

void callMultiStreamingCudaKernel(unsigned char *dev_src, unsigned char *dev_dst, int numChannels, int iw, int ih, cudaStream_t *ptr_stream)
{

    dim3 numOfBlocks((iw / 20), (ih / 20)); //DON'T multiply by 3 because we have 1/3 data of image
    dim3 numOfThreadsPerBlocks(20, 20);
    multiStream_ColorTransformation_kernel << <numOfBlocks, numOfThreadsPerBlocks, 0, *ptr_stream >> >(numChannels, iw, ih, dev_src, dev_dst);

    return;
}

int main()
{

    cudaStream_t stream_one;
    cudaStream_t stream_two;
    cudaStream_t stream_three;

    cudaStreamCreate(&stream_one);
    cudaStreamCreate(&stream_two);
    cudaStreamCreate(&stream_three);

    Mat image = imread("DijSDK_test_image.jpg", 1);
    //Mat image(1080, 1920, CV_8UC3, Scalar(0,0,255));
    size_t numBytes = image.rows * image.cols * 3;
    int numChannels = 3;

    int iw = image.rows;
    int ih = image.cols;
    size_t totalMemSize = numBytes * sizeof(unsigned char);
    size_t oneThirdMemSize = totalMemSize / 3;

    unsigned char *dev_src_1, *dev_src_2, *dev_src_3, *dev_dst_1, *dev_dst_2, *dev_dst_3, *h_src, *h_dst;


    //Allocate memomry at device for SOURCE and DESTINATION and get their pointers
    cudaMalloc((void**)&dev_src_1, (totalMemSize) / 3);
    cudaMalloc((void**)&dev_src_2, (totalMemSize) / 3);
    cudaMalloc((void**)&dev_src_3, (totalMemSize) / 3);
    cudaMalloc((void**)&dev_dst_1, (totalMemSize) / 3);
    cudaMalloc((void**)&dev_dst_2, (totalMemSize) / 3);
    cudaMalloc((void**)&dev_dst_3, (totalMemSize) / 3);

    //Get the processed image 
    Mat org_dijSDK_img(image.rows, image.cols, CV_8UC3, Scalar(0, 0, 255));
    h_dst = org_dijSDK_img.data;

    //while (1)
    for (int i = 0; i < 3; i++)
    {
        std::cout << "\nLoop: " << i;

        //copy new data of image to the host pointer
        h_src = image.data;

        //Copy the source image to the device i.e. GPU
        cudaMemcpyAsync(dev_src_1, h_src, (totalMemSize) / 3, cudaMemcpyHostToDevice, stream_one);
        //KERNEL--stream-1
        callMultiStreamingCudaKernel(dev_src_1, dev_dst_1, numChannels, iw, ih, &stream_one);


        //Copy the source image to the device i.e. GPU
        cudaMemcpyAsync(dev_src_2, h_src + oneThirdMemSize, (totalMemSize) / 3, cudaMemcpyHostToDevice, stream_two);
        //KERNEL--stream-2
        callMultiStreamingCudaKernel(dev_src_2, dev_dst_2, numChannels, iw, ih, &stream_two);

        //Copy the source image to the device i.e. GPU
        cudaMemcpyAsync(dev_src_3, h_src + (2 * oneThirdMemSize), (totalMemSize) / 3, cudaMemcpyHostToDevice, stream_three);
        //KERNEL--stream-3
        callMultiStreamingCudaKernel(dev_src_3, dev_dst_3, numChannels, iw, ih, &stream_three);


        //RESULT copy: GPU to CPU
        cudaMemcpyAsync(h_dst, dev_dst_1, (totalMemSize) / 3, cudaMemcpyDeviceToHost, stream_one);
        cudaMemcpyAsync(h_dst + oneThirdMemSize, dev_dst_2, (totalMemSize) / 3, cudaMemcpyDeviceToHost, stream_two);
        cudaMemcpyAsync(h_dst + (2 * oneThirdMemSize), dev_dst_3, (totalMemSize) / 3, cudaMemcpyDeviceToHost, stream_three);

        // wait for results 
        cudaStreamSynchronize(stream_one);
        cudaStreamSynchronize(stream_two);
        cudaStreamSynchronize(stream_three);

        //Assign the processed data to the display image.
        org_dijSDK_img.data = h_dst;
        //DISPLAY PROCESSED IMAGE           
        imshow("Processed dijSDK image", org_dijSDK_img);
        waitKey(33);
    }

    cudaDeviceReset();
    return 0;
}

UPDATE-1：如果我删除了第一个流的内核调用，那么第二个内核和第三个流的H2D副本会以某种方式重叠（不完全），如下所示。

UPDATE-2 我甚至尝试使用10个流，但事情保持不变。第一个流的内核处理仅在第十个流数据的H2D副本之后才开始。

Answer 1

正如评论者已经指出的那样，主机内存必须是page locked。

无需通过cudaHostAlloc分配额外的主机内存，您可以在现有的OpenCV映像上使用cudaHostRegister：

cudaHostRegister(image.data, totalMemSize, cudaHostRegisterPortable)

为什么cudaMemcpyAsync（主机到设备）和CUDA内核是并行的？

1 个答案: